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TEXT BOOKS FOR ENGINEERS AND STUDENTS. BY DE^ VOLSON WOOD, Late P>'ofessor of Engineering in Stevens Institute of Technohg'^, A TREATISE ON THE RESISTANCE OF MATERIALS, AND AN APPENDIX ON THE PRESERVATION OF TIMBER. Seventh edition. 8vo, cloth $2 00 Tli's work originally consisted chiefly of the Lectures delivered by Professor Wood npon this subject to the classes in Civil Engineering in the University of Michigan, but it has since been twice revised. It is a text-boolc, and gives a brief sketch ot the history of the development of the theories connected with the growth of the subject, and a large amount of experimental matter. Ah English reviewer of the work says : *'It is equal in grade to Rankine's work." THE ELEMENTS OF ANALYTICAL MECHANICS, With numei^ ous examples and illustrations. For use in Scientific Schools and Colleges. Seventh edition, includinp- Fluids. 8vo, cloth $3 00 The Calculus and analytical methods are freely nsed in this work. It contains many problems completely solved, and many others which are left as exercises for the student. The last chapter shows liow to reduce all the equation* of mechanics from the principle of D'AlemberU PRINCIPLES OF ELEMENTARY MECHANICS. Fully illustrated. Tenth edition. 12mo, cloth.' $1 25 The chief aim of this work is to define, explain and enforce the fundamental principles of mechanics. The analysis is simple, but the work is comprehensive. The principles of energy are applied to the solution of many problems. There is a variety of examples at the end of each chapter. A novel feature of the work is the "Exercises," which contain many peculiar and inteiesting questions. SUPPLEMENT AND KEY TO PRINCIPLES OF ELEMENTARY MECHANICS. 12mo, cloth $1 25 This work not only contains a solution of all the examples and answers to the exorcises, out m'ich additional matter which can hardly fail to interest any student of this science. THE ELEMENTS OF CO-ORDINATE GEOMETRY. In Three Parts. I. Cartesian Geometry and Higher Logic. II. Quarternions. III. Modern Geometry and an Appendix.. 1 vol., 8vo, cloth. ^2 00 It was designed to make this a thoroughly practical book for class use. The more abstruse parts of the subject are omitted The properties of the conic sections are, so far . as practicable, treated under common heads, thereby enabling the author lo condense the work. The most elementary principles only of quarternions and modern geometry arc presented ; but in these part^, as well as in the first part, are numerous examples. A TREATISE ON CIVIL ENGINEERING. By Prof. D. H. Mahan. Revised and edited, with additions and new plates, by Prof. De Volson Wood. With an Appendix and.Complete Index. 8vo, cloth. . .$5 00 TRIGONOMETRY. Analytical, Plane, and Spherical. With Logarithmic Tables. Fourth edition. By Prof. De Volson Wood. 12mo, cloth. $1 00 THERMODYNAMICS. By Prof. De Volson Wood, C.E.,M. A. 8vo, cloth. Eighth edition .' $4 00 THEORY OF TURBINES. By Prof . De Volson Wood. 8vo, cloth. $2 50 ELEMENTS or ANALYTICAL MECHANICS SOLIDS AND FLUIDS. BT DeVOlso:n" wood, a.m., c.b., I PBOrBSSOB OF MATHEMATICS AND MBCHANICS IN STEVENS INSTITUTK OK TKCIINOIXKST J AUTHOR OB- "RESISTANCE OF MATERIALS;" "ROOFS AND BRIDGES;" REVISED SDITION OB" " MAHAN'S CIVII. ENGINEBRINa," AND " ELEMENTARY MECHANICS," SEVENTH EDITION. FIFTH THOUSAND.' NEW YORK JOHN WILEY & SONS LONDON CHAPMAN & HALL, Ltd. 1905 ^':>jA / Copyrighted, 1876, by Devolson wood. Renewed 1904, by Frances H. Wood. PRESS OF BRAUNWORTH & CO- BOOKBINDERS AND POINTERS BROOKLYN, N. Y. ^ PREFACE. The plan of this edition is the same as the former one. It is designed especially for students who are beginning the study of Analytical Mechanics, and is preparatory to the higher works upon the same subject, and to Analytical Physics and Astro- nomy. The Calculus is freely used. I have sought to present the subject in such a manner as to familiarize the student with analytical processes. For this reason the solutions of problems have been treated as applications of general formulas. Tlie solution by this method is often more lengthy than by special methods; still, it has advantages over the latter, because it establishes a uniformity in the process. My experience has shown the importance of applying the fundamental equations to a great variety of problems. I have, therefore, in Article 24, and Chapters lY. and X., given a large number and a considerable variety of problems to be solved by the general equations under which they respectively fall. in the revision I have been aided not only by my own expe- rience with the use of the former edition in the class-room, but also by the friendly advice and criticism of several professors of colleges who have used the work. The result has been that several pages have been rewritten, some definitions changed, and the typographical errors corrected. Several new pages in the latter part of the work have been added. I am especially indebted to Professor E. T. Quimby, of Dartmouth College, Hanover, N. H., for his valuable suggestions and for assistance in reading the final proofs. The nature of force remains as much a rayster}- as it was 259817 IV PBBFACB. when its principles were first recognized. Of its essential nature we shall probably remain forever in ignorance. We can only deal with the laws of its action. These laws are deter- mined by observing the effects produced by a force. Force is the cause of an action in the physical world. The results oi the action may be numerous and varied. Thus, force may pro- duce pressure, tension, cohesion, adhesion, motion, affinity, po- larity, electricity, etc. Or, to speak more properly, since force may be transmuted from one state to another, we would say that the above terms are names for the different manifestations of force. The question " what is the correct measure of force " has taken different phases at different times. During the last century it was contended by some that momentum {Mv) was the correct measure, while others contended that it should be the work which it can do in a unit of time {^Mv^). But as one has happily expressed it, '" theirs was only a war of words ; " for the real measure of force enters only as a factor in the expressions. Thus, if i^ be a constant force, the value of the momentum is Ft, see page 51, and of the work Fs^ see page 45. At the pre- sent day some contend that the only measure of force is the motion which it produces, or would produce, in a unit of time. This is called the absolute measure, and the absolute unit of FOKCE is the velooity which the force jproduces^ or would jpro- duce^in a unit of mass in a unit of time if it acted during the unit with the intensity which it had at the instant con- sidered. If the intensity of the force were constant, the velo- , city which it produced at the end of the unit of time would be the required velocity. Hence, the absolute measure of any force acting on any mass is the product of the mass into the acceleration; and is the second member of equation (21). This is a correct measure, and is accepted as such by all writers on mechanics. But those who contend that this is the only measure, neces- sarily deny that weighty or more generally jpounds, is a mea- snre. I contend that pounds is a measure of the intensity of a PKBFACB. V force both statically and dynamically. Many authors maintain the same position. Indeed, it is probable that the position which I have taken can be deduced from any standard work on mechanics ; but in some it is left to inference. Thus, in Smith's Mechanics, P^g® 1> ^ve find this terse and correct defi- nition, " Tlie iritensity of a force may be measured, statically, by the pressure it will produce ; dynamically, by the quantity of motion it will produce." I say this is correct, but 1 will add that the intensity of a force which produces a given motion is also measured by a pressure, or by sometliing equivalent to a pressure, or to a pull. To those who will look at it analyti- cally, it is only necessary to say that the first member of equa- tion (21) is measured in pounds. If we know the absolute measure, we may easily find its value in iwunds. The j>ou'iid here referred to is the result of the action of gra- vity upon a certain quantity of matter. The amount of matter having been fixed, either by a legal enactment or by common consent, and declared to be one pound at a certain place, its weight, as determined by a standai'd spring-balance at any other place, becomes a measure of the force of gravity as compared with the fixed place. This standard spring-balance may meas- ure the intensity in pounds of any other force, whether the body upon which the force acts be at rest or in motion. If a perfectly free body were placed in a hollow space at the centre of the earth, at which place it would be devoid of weight, and pulled or pushed by a constant force, whose intensity, measured by a standard spring-balance, equaled the weight of an equal body on the surface of the earth, then would its motion be the game as that of a falling body. See page 24, Problem 7. In the forces of nature producing motion, there being no visible coimection between the point of action of the force and the body upon wliich it act&, we are unable to wei(//i their intensity except by calculation. If the absolute measure is known, the pounds of intensity may be computed. The absolute measure of the force of gravity on a mass m is mg, and the weight of tlie body being W, we have W= mg. The sun acts upon the earth with a force which may be expressed by the absolute VI PEEFAOB. measure, and also by a certain number of pounds of force. More than half of the examples in Article 24 involve ao equality between j[)ounds of intensity and the absolute measure of the force. The tact is, that, in case of motion, these quanti- ties are co- relative. Since, then, it is correct to use the term jpound as the measure of the intensity of a force whether the body be at rest or in motion, and since it is in common use, and the student is familiar with it, I prefer to consider a force as measured by a certain number of pounds. See Article 9. It is more simple, containing as it does only one element, than the absolute measure, which contains three elements — mass, velocity, and time. There is another advantage in thus measuring force. Stu- dents frequently, and in some cases writers, use the expressions, *' quantity of force," " amount of force," " force of a blow," etc., when they mean (or should mean) momentum, or work, or via viva. lu such cases an attempt to answer the question "how many pounds of force " would show at once that the quantity re- ferred to was wot force. So much ambiguity, or at least indeiiniteness, has arisen in regard to the term force, that I have rejected the terms " Im- pulsive Force " and " Instantaneous Force," and used the term " Impulse " instead of them. We know nothing of an instan- taneous force, that is, one which requires no time for its action. I also reject the expression y6>rc6 of inertia. I do not believe that inertia is 2, force. To the question " The inertia of a body is how many pounds of force " there is no answer. The term moment of inertia has no physical representation- The nearest approach to it is in the expression for the vis viva of a rotating body. In such problems the moment of inertia forms an important factor. The energy of a rotating body hav- ing a constant angular velocity is directly proportional to its moment of inertia in reference to its axis of rotation. See page 202. But motion is not necessary for its existence. See page 165. The expression appear? in the discussion of numerous PREFACE. yU •statical problems, such as the flexure of a beam, the centre of pressure of a fluid, the centre of gravity of certain solids, etc. It is not the moment of a moment, although it may be so con- strued as to appear to be of iimtjbrm. Some other term might be more appropriate. Even the expression moment of the mass would be less objectionable. The subjects of Centrifugal Force and Unbalanced Force have been discussed of late in Engineering. Some assert that there is no such thing as a centrifugal force. Much unprofita- ble discussion may be avoided by strictly defining the terms used. If it is defined to be a force equal and opposite to the deflecting force, it will, at least, have an ideal existence, just as the resultant in statical problems has an ideal existence. But the vital question is, is the centrifugal force active when the deflecting force acts ? Or, in other words, do both act upon the body at the same time ? It seems, however, quite evident that if both acted upon the body at the same time they would neutralize each other, and the body would move in a straight line. Hence, in the movement of the planets, or of any free rotating body, there is no centrifugal force. But in the case of a locomotive running around a curve there may be both cen- tripetal and centrifugal forces ; the former acting against the locomotive to force it away from a tangent to tlie track ; the latter, against the track, tending to force it outward. Wher- ever the force is conceived to act, whether just between the rail and wheel or at some other point, it is evident that both do not act upon the same body. Similarly in regard to the unbalanced force. It is a conveni «nt term to use, but, in a strict sense, an unbalanced force does not exist ; for action and reaction are equal and opposite. But in reference to a particular body, all other conditions being ignored, the force may be unbalanced. Thus, when a ball is fired from a cannon, the force of the powder, considered in the direction of the motion of the ball only, is unbalanced ; but the powder exerts an equal force in the opposite direction, and in that sense also is unbalanced. But when the entire effect of the Vlll PREFACE. force in all directions is considered, the algebraic resultant is zero. In other words, the centre of gravity of the system, for forces acting between its integrant parts, remains constant. These are some of the fundamental questions which will arise in the mind of the student as he studies the subject. Fortu- nately, it is not necessary for him to settle them beyond the question of a doubt before he proceeds with the subject. On some of these points scholars, who have made the subject a specialty, differ ; and it is only after a careful consideration of the points involved that one can take an intelligent position in regard to them. DbYolson Wood. HflBOKSK, AytgutL, 1837. GREEK ALPHABET. Letters. A a Names. Alpha Beta Lettera N V Hamea. Nn Xi r^ Gamma Omicron A B Delta n TT Pi E € Epsilon Zeta Eta 5 O- 9 T T Kho Sigma Tau se Theta T V Upsilon I I Iota $<p Phi K K A \ Kappa Lambda Xx Chi Psi M a Mu n a> Omega TABLE OF CONTENTS. CHAPTER L DXFINmONB AND THE LAWS OF FORCES WHICH ACT ALONG A 8TBAIGHT LINB. .AKTICLB8 VA«a 1-14 — Definitions. 1 15-20— Velocity ; Gravity; Maes : 6 21-23— Force; Mass; Density 16 24 — Examples of Accelerated Motion 21 25-27— Work ; Energy ; Momentum 44 28-33— Impact 53 84r-36— Statics ; Power; Inertia 68 87-38 — Newton's Laws of Motion. Eccentric Impact 88 CHAPTER n. COMPOSITION AND RESOLUTION OF FOKCBS. 89-42— Conspiring Forces 66 48-46 — Composition of Forces 65 47 — Resolution on Three Axes 70 48-49 — Constrained Equilibrium of a Particle 71 60-60— Statical Moments 79 Examples 84 CHAPTER IIL PARALLEL FORCE& 61-63— Resultant 86 64— Moments of Parallel Forces 8^ 65-68— Statical Couples 88 69-74— Centre of Gravity of Bodies 93 75— Centrobaric Method 103 77-81 — General Properties of the Centre of Gravity 108 82~Centre of Mass UO Xll TABLE OF CONTENTS. CHAPTER IV. HOKCONCUBRBNT FOBCB& ABVIOLBB KAOS 83- 85— General Equationfl. HI 86— Problems llfr CHAPTER V. STRESSES. 87- 90— Stresses Resolved 143 91- 92 — Shearing Stresses — Notation 144 93_ 94— Resultant Stress • 140 95- 96 — Discussion 158 97- 98 — Conjugate Stresses. General Problem 156 CHAPTER VL VIRTUAL VELOCITIES. 100 — Concurring Forces 159 101 — ^Noncomjurring Forces 160 Examples 161 CHAPTER VIL MOMENT OP INERTIA. 103-104— Examples. Formula of Reduction 165 105 — Relation between Rectangular Moments 169 106— Moments of Inertia of Solids 173 107— Radius of Gyration. ; 178 CHAPTER Vni. MOTION OF A FREE PARTICLE. 108— General Equations 176 109 — ^Velocity and Living Force 176 110-117— Central Forces 189 CHAPTER IX. CONSTRAINED MOTION OF A FABTICUB. 118-131— General Equations Idl 133-134— Centrifugal Force on the Earth 188 TABLE OF CONTENTS. Xlll CHAPTER X. BOTATION OF A BODY WHEN THE FOB0E8 ABB Df A PLAIVB. AsnoLM 9Amm 126-131— General Equationfl 200 132-Reduced Mass 204 188-135 — Spontaneotis A-ria ; Centre of Percussion ; Axis of Instantaneous Rotation 209 Examples 210 18^138 — Compound Pendulum. Captain Eater's Experiments. 318 189-142— EUipticity of the Earth. Torsion Pendulum. 221 143— Density of the Earth 224 144_146_Problem8 227 CHAPTER XI. QENBRAL EQUATIONS OF MOTIOH. 147— D'Alembert's Principle 230 14S_General Equations 231 149 — Conservation of the Centre of Gravity r . . . 235 150 — Conservation of Areas 235 151 — Conservation of Energy, 236 152 — Composition of Angular Velocities 240 153 — ^Moments of Rotation of the Centre 242 154— Motion of the Centre of a Body 242 155— Motion of Rotation of the Centre 243 156— Euler's Equations .243 157 — Principal Axes 244 158 — No Strain on Principal Axes 246 159 — Relation between the Fixed and Principal AxeB 247 160— Relations between Angular Velocities 247 161 — ^Vxis of Spontaneous Rotation 253 163 — Relation between Spontaneous and Central Axes 255 164 — Centre of Percussion 263 165 — Conservation of the Centre of the Mass 264 166 — Conservation of the Moment of the Momentum 264 167— The Invariable Plane 266 168 — Mutual Action between Particles 2G7 169 — Attraction of a Homogeneous Sphere 268 169a-Mass and Stress 271 1696-Repulsive Forces 2'J^ 170— Principle of Least Action 273 170a-Gauss' Theorem of Least Constraint 274 XIV TABLE OF CONTENTS. CHAPTER Xn. MECHANICS OF FLXnDS. ABTICLES TAOn 171 — Introductory 275 173— Definitions 276 173— Elastic and Non-Elastic Fluids 376 174 — Equality of Pressures 276 175— Perfect Fluid at Rest 277 176— The Law of Equal Transmission 277 177 — Pressure on Base of Prismatic Vessel 278 178— Pressure on Base of any Vessel 278 179— Static Head 279 180— Free Surface 279 181 — Pressure on Submerged Surface 280 182— Resolved Pressures 281 183— Resultant Pressure 281 184— Centre of Pressure 282 185— Condition of Flotution 285 186— Depth of Flotation 286 187— Specific Gravity .288 188— Examples 289 189— Fluid Motion. Definitions 291 190— Bernouilli's Theorem 292 191 -Discussion of 294 192 — Graphic Representation 295 193 — Flowing through Varying Sections. Examples 296 194— Velocity of Discharge from Small Orifice 300 195 — Discharge from Large Orifice .... 301 196— Vertical Orifices. Examples 302 197— Conical Adjutages 307 198— Reaction of Fluids .310 199— Centrifugal Force 310 200— Resolved Centrifugal Force 311 202 — Pressure Due to Discharge from the Side of a Vessel 314 203 — Pressure Due to a Discharge Vertically Upward 315 205 — Pressure of a Stream Impinging Normally Against a Plane Sur- face 316 206— Case of Cup Vane 316 207— Case of Bent Tube 317 208— Case of Inclined Surface ^319 209— Remark. Examples ; 319 210— Case of Impinged Surface in Motion 322 211— Work Done on Vane 333 212— Energy Imparted to a Vane 324 TABLE OF CONTENTS. XT JATICLKR PAQB 213— Expression for Efficiency 825 214— Resultant Pressure on a Vane 82S 215— General Case 326 216— Discussion 327 217— Hydraulic Motors 832 218— Case of a Single Vane 382 220— The Undershot Wheel 383 222— The Poncelet Wheel 836 223— The Breast Wheel 887 224— The Overshot Wheel. Examples 337 225— Effective Head 342 226— The Jet Propeller 343 227— Barker's Mill 343 228 — Pressure Due to Centrifugal Force. Examples 348 229— Definitions of Turbines 350 230— Analysis of Turbines 351 231— Special Cases 855 282— General Case. Examples 859 233 — Resistance Due to Sudden Enlargements 862 234 — Resistance in Long Pipes 863 235 — General Case of Flow in Long Pipes, Examples 864 236— Gases 369 237— Boyle's Law 370 238— Tension of Gas in a Vertical Prism 371 289— Numerical Values in Regard to Air 372 240— Tension of the Atmosphere Gravity Variable 372 241— Heights by Barometer 374 242 — Gay Lussac's Law 375 243— Absolute Zero 376 244— Heat a Form of Energy 377 245— Thermal Unit 378 246— Specific Heat 878 247— Dynamic Specific Heat 379 248— The Adiabatic Curve. Examples 379 249^— Velocity of a Wave in an Elastic Medium 382 250— Value of ;' 387 251— Remarks. Examples 387 252— Velocity of Discharge of Gases 388 254— Weight of Gas Discharged 390 APPENDIX I. Solutions of Problems 391-463 APPENDIX II. The Potential 467-470 ANALYTICAL MECHANICS. CHAPTER L DKFINITIONS, AND PRIN0IPLE8 OF ACTION OF A SINGLE FOBOB, AUD OF FORCES ACTING ALOJTG THE SAME LINE. 1. Mechanics treats of the laws of forces, and the equi- librium and motion of bodies under the action of forces. It has two grand divisions, Dynamics and Statics. 2. Dynamics treats of the motion of material bodies, and the laws of the forces which govern their motion. 3. Statics treats of the conditions of the equilibrium of bodies under the action of forces. There are many subdivisions of the subject, such as Hydrodynamics, Hy- drostatics, Pneumatics, Thermodynamics, Molecular Mechanics, etc. That part of mechanics which treats of the relative motion' of bodies which are so connected that one drives the other, such as wheels, pulleys, links, etc., in machinery, is called Cinematics. The motion in this case is independent of the intensity of the force which produces the motion. Theoretic Mechanics treats of the effect of forces applied to material points or particles regarded as without weight or magnitude. Somatology is the application of theoretic mechanics to bodies of definite form and magnitude. 4. Matter is that which receives and transmits force. lu a physical sense it possesses extension, divisibility, and impene- trability. Matter is not confined to the gross materials which we see and handle, but includes those substances bj which sound, heat, light, and electricity are transmitted. It is unnecessary in this connection to consider those refined speculations by which it is sought to determine the essential nature of matter. According to ■ome of these speculations, matter does not exist, bat is only a oonoeption. 2 DEFINITIONS. [5-6.) According to this view, bodies are forces, within the limit of which the attrac- ti-ve exceed the repulsive ones, and at the limits of which they are equal to each other. But observation, long continued, teaches practically that matter is inert, that it has no power within itself to change its condition in regard to rest or motion ; that when in motion it cannot change its rate of motion, nor be brought to rest without an external cause, and this cause we call FORCE. One also learns from observation that matter will transmit a force, as for instance a pull applied at one end of a bar or rope is transmitted to the othei end ; also a moving body carries the effect of a force from one place to another. 5. A Body is a definite portion of matter. A particle is an infinitesimal portion of a body, and is treated geometrically as a point. A molecule is composed of several particles. An atom is an indivisible particle. 6. Force is that which tends to change the state of a body in regard to rest or motion. It moves or tends to move a body, or change its rate of motion. We know nothing of the essential nature of force. We deal only with the laws of its action. These laws are deduced by observations upon the effects of forces, and on the hypothesis that action and reaction are equal and ojpjposite / or, in other words, that the effect equals the cause. In this way we find that forces have different intensities, and that a relation may be established between them. It is necessary, therefore, to establish a unit. This may be assumed as the effect of any known force, or a multiple part thereof. The effect of all known forces is to produce a pull, or push, or their equivalents, and may be measured by pounds, or by something equivalent. The force of gravity causes the weight of bodies, and this is measured by pounds. We therefore assume that a standard POUND is the UNIT of force. The standard pound is established by a legal enactment, and has been so fixed that a cubic foot of distilled watei* at the level of the sea, at latitude 45 degrees, at a temperature of 62 degrees Fahrenheit, with the barometer at 30 inches, will weigh about 62.4 pounds avoirdupois. The English standard pound was originally 5,760 troy grains. The grain was the weight of a certain piece of brass which was deposited with the clerk of t-he House of Commons. This was destroyed at the time of the burning of the House of Commons in 18eS4, after which it was decided that the legal [7.] STANDARD UNITS. 3 pound should be the weight of a certain piece of platinum, wei^ning 7,000 grains. This is known as the avoirdupois pound, and the troy pound ceased to be the legal standard, although both have remained in common use. The legal standard pound in the United States is a copy of the English troy I)Ound, and was deposited in the United States Mint in Philadelphia, in 1827, where it has remained. The avoirdupois pound, or 7,000 grains, is used in nearly all commercial transactions. The troy pound is a standard at G2 de- grees Fahrenheit and 30 inches of the barometer. The weight of a cubic inch of water at its maximum density, as accepted by the Bureau of Weights and Measures of the United States, is stated by Mr Hosier, in a report to the Secretary of the Treasury. 1842, to be 252.745b grains. Mr. Hasler determined the temperature at which water has a maxi. mum density, at 39.83 degrees Fahrenheit, but Playfair and Joule determined it to be 39.101" F. The exact determination of the equivalent values of the units is very diffi • cult, and has been the subject of much scientific investigation. — (See The Me- tric System, by F. A. P. Barnard, LL.D., New York, 1872.) When a quantity can be measured directly, the unit is generally of the same quality as the thing to be measured : thus, the unit of time is time, as a day or second ; the unit of length is length, as one inch, foot, yard, or metre ; the unit of volume is volume, as one cubic foot ; the unit of money is money ; of weight is weight ; of momentum is momentum ; of work is work, etc. When dissimilar quantities are used to measure each other a proportion must be established between them. It is commonly said that "the arc mea- sures the angle at the centre," but it does not do it directly, since there is nc ratio between them. The arc is a linear quantity, as feet or yards, or a num- ber of times the radius, while the angle is the divergence of two lines, and ia usually expressed in degrees. But angles are proportional to their subtended arcs ; hence we have an equality of ratios, or angle subtended a/re unit angle ~ are which subtends the unit angle* and since a semi-circumference, or «-, subtends an angle of 180°, it is easy from the above equality of ratios to determine any angle when the arc is known, or vice versd. Similarly, the intensity of heat is not measured directly, but by its effect in expanding liquids or metals. The magnetic force is measured by its effect upon a magnetic needle. The intensities of lights by the relative shadows produced by them. Similarly with forces, we measure them by their effects. Di8"9imilar quantities, between which no proportion exists, do not measim each other. Thus feet do not measure time, nor money weight. Pounds for commercial purposes represents quantities of matter ; but when applied to forces it represents their intensities. In a strict sense, pounds does not measure directly the quantity of matter, but is always a measure of a force. 7. The line of action of a force is the line along which the force moves or tends to move a particle. If the particle is 4 FORCE, SPACE, TIME. [8-141 acted tipon by a single force, the line of action is straight This is also called the action-line of the force. 8. The poiot^ of application of a force is the point at which it acts. This maj be considered as at any point of its action-line. Thus, if a pull be applied at one end of a cord, the effect at the other end is the same as if applied at any intermediate point. 9. A FOKCE is said to be given when the following elements are known : — 1st. Its magnitude {pounds) ; 2d. The direction of the line along which it acts (action-line) ; 3d. The direction along the action- line (+ or — ); and, 4:th. Its point of application. A force may be definitely represented by a straight line ; thus, its magnitude may be represented by the length A B, Fig. 1 ; its position by the A ^ B position of the line A B -, its direction along ^^^ , the line by the arrow-head at B, which indi- cates that the force acts from A towards B ; and its point of application by the end A. 10. Space is indefinite extension, finite portions of which may be measured. 11. Time is duration, and may be measured. Probably no definition will give a better idea of the abstract quantities of time and 8pa4ie than that which is formed from experience. 12. A BODY is in motion w^hen it occupies successive portions of space in successive instants of time. In all other cases it is at absolute rest. Motion in reference to another moving body is relative. But a body may be at rest in reference to surrounding ob- jects and yet be in motion. Thus, many objects on the surface of the earth, such as rocks, trees, etc., may be at rest in refer- ence to objects around them, while they move with the earth through space. Observation teaches that there is probably no body at absolute rest in the universe. 13. Motion is uniform when the body passes over ecjual portions of space in equal successive portions of time. 14. Yabiable motion is that in which the body passes over unequal portions of space in equal times. fl5.J VELOCITY. 5 15. Yelocity is the rate of motion. When the motion i9 nniform it is measured by the linear distance over which a body would pass in a unit of time ; and when it is variable it is the distance over which it would pass if it moved with the rate which it had at the instant considered. The path of a moving particle is the Hue which it generates. For uniform velocity, we have .=i-. (1) ' ID which 8 = the space passed over ; t = the time occupied in moving over the space s ; and V = the velocity. For vci/riahle velocity^ we have -I- (^) Examples. 1. If a particle moves uniformly thirty feet in three secondfl, what is its velocity % 2. If 5 = at^ what is the velocity \ 3. \i 8 — a^ -\- ht, what is the velocity at the time t^ or at the end of the space 8 ? Here at which is the answer to the first part. Find t from the given equation, and substitute in the expression for v, and it gives the answer to the second part ; or 'y= Vh^ + 4: as. 4. If 8 = 4^, required the velocity at the end of five seconds. 5. If 3.9* = 5f, required the velocity at the end of ten sec- onds. 6. If 5 = igf, what is the velocity in terms of the time and space ? 7. If a^ = ^~1, required the velocity in terms of the time and space. 6 ANGULAR VELOCITY. [16, 17, J 16. Angular velocity is the rate of angnhir movement. If a particle moves aroand a point having either a constant or a variable velocity along its path, the angular velocity is meas- ured hy the arc at a unites dis- tance which subtends the angle swept over in a unit of time hy ^^^ that radius vector which passes p^^ 2. through the particle. \i s — AB = the length of the path described ; V = the velocity along the path AB ; t = the time of tlie movement ; r = OB = the radius vector ; d = the circular arc at a unit's distance which subtends the angle A OB swept over by the radius vector in the time t ; and 6> = the angular velocity ; Then, if tlie angular motion is uniform, (8) If it is variable^ then t dd We also have, ds = vdt= \/^iH¥~Vd^\ jm ds^ — dr^ "^^df — d?^ , ••■ '^ = — 75- - = — ^ — ; an<i d0 VM|)' (6) '•'^Jt=' r 17, Acceleration is the rate of increase or decrease of the velocity. It is a velocity-increment. The velocity-increment of an increasing velocity is considered positive, and that of a decreasing velocity, negative. The measure of the acceleration, when it is uniform, is the amount by which the velocity is increased (or decreased) in a T17.] ACCELERATION. 7 onit of time. If the acceleration is vaiiable, it is the amoiiLt by which the velocity would be increased in a unit of time, provided the rate of increase continued the same that it was at the instant considered. Hence, if y = the measure of the acceleration (ci briefly, the acceleration) ; then, when the acceleration is uniform, • d8_ „ V dt and hence, when it is variable, '^~dt~ dt^W ^^ We also have ^_ dh ds^ _ d''s dsi _ ^d^8 ,g,x ^~~de^ dS'~d?^~dt^'~ W ^^ We thus see that the relation between space, time, and velo- city are independent of the cause which produces the velocity. Applications of Equation (6). 1. Suppose that the acceleration is constant. Then in (6) y will be constant, and dt being the equicrescent variable, we have 4''' = itf'''\ ory^ = g+C.. (7) ds But for ^ = 0, -3- = Vo = the initial velocity .'. 61 = — i\, ; and (7) becomes ds =ftdt + Vffitt integrating again gives « = -i/i? + v^t + (7a. S EXAMPLES. [IT.] But 8 = So for ^ = .*. 6i = Sqi hence the final equation is 8 = iff + Vo6-^8ol (8) which gives the relation between the space and time. Again, multiplying both members of equation (6) by dt, we have ^fds<Ps=/fd8; or _=,2/^ + ^o'; (9) ds ,\ dt = VV + P' and integrating, gives t = '^J^''+a (10) Equation (7) gives the relation between the velocity and time, and equation (9) between the velocity and space. If Vq and «o are both zero, the preceding equations become 2« V =zft = ^2/5 = -7-. (11) 8=^iff = ^^=^t (12) t = -.= s/-^ = -, (13) We shall find hereafter that these formulas are applicable to all cases in which the force is constant and uniform. 2. Find the relation between the space and time when the acceleration is naught. We have df-^' Multiply by dt, integrate twice, and we have 5 = 5o + Vo^ ; in which .% and v^ are initial values; that is, the body will have [17.] EXAMPLES. 9 passed over a space «© before t is computed. Vo is not only the initial bnt the constant uniform velocity. If «© = 0, then * = V. 8. If the acceleration varies directly as the time from a state of rest, required the velocity and space at the end of the time t. Herey= at. 4. Determine the velocity when the acceleration varies in- versely as the distance from the origin and is negative ; OTf=r a 8 6. Determine the relation between the space and time when the acceleration is negative and varies directly as ,the distance from the origin ; OTf= — hs. Equation (6) becomes Multiplying both members by ds^ we have dsd^a de Integrating gives = — h9d»> But v = for « = j^ /. Gx — W; and d^ or h^ dt = Integrating again gives ^*r = sin-^ -+ ft. But ^ = for * = «^ .*. ft = — J^ ; .-. « = «o sin {tb^ + W- (15) 10 ACCELERATION. PT.I « 5 = 0, 2J = ^ttJ-*, IttJ-*, f-TrS -♦, JttJ-* |7r5-», etc.; = — So, ^ = ttJ *, SttJ-S SttJ-*. This is an example of periodic motion, of which we shall have examples hereafter. 6. Determine the space when the acceleration diminishes as the square of the velocity. When the acceleration is constant, the relation between the time, space, and velocity may be shown by a triangle, as in Fig. 3. Let A^ represent the time, say four seconds. Divide it into four equal spaces, and each space will represent a second. Draw horizontal lines through the points of division and limit them by the inclined line AC The horizontal lines will represent the corresponding velocities. Thus V2 = ge is the velocity at the end of the time 4* The triangle Aho represents Fig- 3. the space passed over during the first sec- ond, and ABO the space passed over during four seconds. The lines de^ fh^ and ^ 6^ represent the accelerations for each second, which in this case are equal to each other, and equal to hOy which is the velocity at the end of the first second. Hence, when the acceleration is uniform, the velocity at the end of the first second equals the acceleration. This is also shown by Eq. (11) ; for if t =l^v =f Equations (12) and (13) may be deduced di- rectly from the figure. If acceleration constantly varies, the case may be represented as in Fig. 4. To find the acceleration at the end of the first second, draw a tangent ae to the curve at the point <z, and drop the perpendicular ad^ then will de be the acceleration. But — , = -^=^ = /*= the ad ab at *^ velocity-increment, which is the same as Equation (6). Fig. 4. [18.] RESOLVED VELOCITIES AND ACCELERATIONS. K 18. Resolved velocities and accelerations. When the motion is along a known path and at a known rate, the projec- tions of the velocities and accelerations upon other paths which are inclined to the given one will equal the product of these quantities bj the cosine cf the angle between the paths ; that is, v' ^= V cos d =1 -J COS 6, andy = -^ cos 6, dt df where v' and/*' are on the new path, and 6 the angle between the paths. Examples. 1. If the velocity v is constant and along the line AB, which makes an angle 6 with the line AC, ' then will the velocity projected on A C also be constant, and equal to v cos 6 ; and on the line BC, equal to V sin d. ^»«- 5. 2. Let ABGhQ a parabola whose equation is ^ = 2^. If a body describes the arc -5 6^ with such a varying velocity that its pro- jection on BD^ a tangent at B, is constant, required the velocity and the acceleration parallel to BE, From the equation of the curve we have dy dx y' From the conditions of the problem we have -jT = constant = v ; dt ' but dx ^dy dx ^ y ^ ^ /2a? which is the velocity parallel to x ; 12 EXAMPLES. ria> d^x _v' dy _ v^ "d^~'^ di^'p ' hence the acceleration parallel to the axis of x will be constant. Let cfe be an element of the arc, then will the velocity along Jie arc be ds dt 3. Determine the accelerations parallel to the co-ordinate axes X and y, so that a particle may describe the a/rc of a parabola with a constant velocity. Let the equation of the parabola be ^ dy_jp_ " dx y' The conditions of the problem give ds -j\ = constant = v, at ^^^dt" dt'dy~dt dy "dt^^^dy" -dt^ 1+1=., ^dy jpv " dt " \ ^f^r g, gives <% jpmj dy d^ ~ {f + f)^ dt fv^y {f + ff which being negative shows that the acceleration perpendicular to the axis of the parabola constantly diminishes. Similarly we find d^x jpV T18.1 EXAMPLES. 13 4. A wheel rolls along a straight line with a uniform velocity; compare the velocity of any point in the circumference with that of the centre. Fio.7. Let V = the velocity of any point in the circumference, v'= the uniform velocity of the centre, r = the radius of the circle, X = the abscissa which coincides with the line On which it rolls, and y = the ordinate to any point of the cycloid. Take the origin at A. The centre of the circle moves at the same rate as the successive points of contact ^. The centre is vertically over B. The abscissa of the point of contact cor- responding to any ordinate y of the cycloid, is r versui-^ -; ' r dy^ dy " dt -T. = ^' y<^ry-f The equation of the cycloid is aj = /• versin - — (2ry — y*)* ; and from the theory of curves d^ = da^ + df.\ ^_ / da? ds dy ds _ / 2r ^ ~ V Qr^TT/' dy y and, _ds _dyd8_/2y ^"di-dt dy-yT'^ 14 GRAVITY. [19.] If y^O, v = 0- y = r, V = \^W ; y =z ^^ V =i v' . Hence, at the instant that any pomt of the wheel is in con« tact with the straight line, it has no velocity, and the velocity at the highest point is twice that of the centre. The velocity at any point of the cycloid is the same as if the wheel revolved ahout the point of contact, and with the same angular velocity as that of the generating circle. For, the length of the chord which corresponds to the ordi- nate 2/ is -v/ 2ry, and hence, if V : 2v' : : V2ry : 2r ; /2y we have v —\/ — ^' as before found. 19. Gravitation is that natural force which mutually draws two bodies towards each other. It is supposed to extend to every particle throughout the universe according to fixed laws. The force of gravity above the surface of the earth diminishes as the square of the distance from the centre increases, but within the surface it varies directly as the distance from the centre. If a body were elevated one mile above the surface of the earth it would lose nearly -guVir of its weight, which is so small a quantity that we may consider the force of gravity for small elevations above the surface of the earth as practically constant. But it is variable for different points on the surface, being least at the equator, and gradually increasing as the latitude increases, according to a law which is approxi- mately expressed by the formula g = 32.1726 - 0.08238 cos 2Z, in which Z = the latitude of the place, g = the acceleration due to gravity at the latitude Z, or simply the force of gravity, and 32.1726 ft. = the force of gravity at latitude 45 degrees. [19.] EXAMPLBS. 15 From this we find that at the equator g = g^ = 32.09022 feet, and at the poles g = ggQ = 32.25498 feet. The varying force of gravity is determined by means of a pendulum, as will be shown hereafter. It is impossible to de- termine the exact law of relation between the force of gravity at different points on the surface of the earth, for it is not homogeneous nor an exact ellipsoid of revolution. The delicate observations made with the pendulum show that any assumed formula is subject to a small error. (See Mecanique CSleste, and Puissanfs Geodesie,) Substituting g for /'in equations (11), (12), and (13), we have the following equations, which are applicable to bodies falling freely in vacuo : — > 2^ v = gt= V2gs = j ; 8 = igf=^^ = ivt; > (16) — ^ — /?f — ?? Examples. 1. A body falls through a height of 200 feet ; required the time of descent and the acquired velocity. Let g = 32^ feet, Ans. ^ = 3.53 seconds. V = 113.31 feet. 2. A body is projected upward with a velocity of 1000 feet per second ; required the height of ascent when it is brought to rest by the force of gravity, X Ans. 8 = 15,544 feet, nearly. 3. A body is dropped into a well and four seconds afterwards it is heard to strike the bottom. Required the depth, the velocity of sound being 1130 feet per second. Ans. 231 feet. 4. A body is projected upward with a velocity of 100 feet per second, and at the same instant another body is let fall from a height 400 feet above the other body ; at what point will they meet? 16 DYNAMIC MEASURE OF A FORCE. [20, 21.1 5. With what velocity mnst a body be projected downward that it may in t seconds overtake another body which has already fallen through a feet f a 1 Am. T = -r+ V2ag. 6. Required the space passed over by a falling body during the n^^ second. 20. Mass is quanttit of matter. If we conceive that a quantity of matter, say a cubic foot of water, earth, stone, or other substance, is transported from place to place, without expansion or contraction, the quantity will remain the same, while its weight may constantly vary. If placed at the centre of the earth it will weigh nothing ; if on the moon it will weigh less than on the earth, if on the sun it will weigh more; and if at any place in the universe its weight will be directly as the attractive force of gravity, and since the acceleration i? also directly as the force of gravity, we have — = constant, for the same mass at all places. This ratio for any contem- poraneous values of W and g may be taken as the measure of the mass, as will be shown in the two following articles. The weight in these cases must be determined by a spring balance or its equivalent. 21. Dynamto measure of a force. Conceive that a body is perfectly free to move in the direction of the applied force, and that a constant uniform force, which acts either as a puU or jpushy is applied to the body. It will at the end of one second produce a certain velocity, which call -^(1,. If now forces of different intensities be applied to the same body they will produce velocities in the same time which are proportional to the forces ; or foe V^x)y in which f is the applied force. Again, if the same forces are applied to bodies having differ- ent masses, producing the same velocities in one second, then will the forces vary directly as the masses, or, f oc Jf . ^21.] DYNAMIC MEASURR OP A FORCE. 17 Hence, generally, if uniform^ constant forces are applied to different masses producing velocities v^, in one second, then f Qc J/^(i, ; or, in fhe form of an equation, f = ^i^%,; (17) where c is a constant to be determined. If the forces are constantly varying^ the velocities generated at the end of one second will not measure the intensities at any instant, but according to the above reasoning, the rate ofvarior tion of the velocity will be one of the elements of the measure of the force. Hence if F z= Q, variable force ; M = the mass moved ; -^=f= the rate of variation of the velocity; or velocity-increment ; and, -^ be substituted for Vju in equation (17), reducing by equation (6), we have fjrt) d^^ F= cMf^ cM^^ = cM^,. (18) From this we have F cM — -Ti; hence the value of cM is expressed in terms of the constant ratio of the force F to that of the acceler- ation yi A To determine this ratio experimentally I | suspended a weight, PT, by a very long fine wire. The wire should be long, so that the body will move practically in a straight line for any arc through which it will be made to move, and it should be very small, «o that it will contain but little mass. By ^ means of suitable mechanism I caused a viq.s. constant force, F, to be applied horizontally to the body, thus causing it to move sidewise, and determined 18 UOTT OF MASS. [22.> the space over which it passed during the first second. This equalled one-half the acceleration (see the first of equations (12) when t — 1). I found when F= ^^TF, that /= 1.6 feet, Clearly ; and for i^= y^ ^> /= ^-^ feet, nearly ; and similarly for other forces ; hence gM = -^ TF, nearly. But the ratio of F to f is determined most accurately and conveniently by means of falling bodies; for ^/=: ^ = the acceleration due to the force of gravity, and W the weight of the body (which is a measure of the statical effect of the force of gravity upon the body), hence W cM= — ', (19) in which the values of W and g must be determined at the same place ; but that place may be anywhere in the universe. The value of c is assumed, or the relation between c and M fixed arbitrarily. If (? = 1 , we have W Jf=y; (20) and this is the expression for the mass, which is nearly if not quite universally adopted. This in (18) gives ^ ^^^5 W(Ps and hence the dynamic measure of the pbessubb whioh moves A body is the product of the mass into the acceleration. This is sometimes called an accelerating force. If there are retarding forces, such as friction, resistance of the air or water, or forces pulling in the opposite direction ; then the first member F, is the measure of the unbalanced forces in pounds, and the second member is its dynamic equivalent 22. Unit of Mass. If it is assumed that c == 1, as in' the preceding article, the unit of mass is virtually fixed. In (20) if 1F= 1 and p' = 1, then JI/'= 1 ; that is, a unit of tyulss is the quantity of matter which will weigh one pound at that \22,i UNIT OP MASS. 19 place in the universe where the acceleration dne to gravity ia one. If a quantity of matter weighs 32^ lbs. at a place where ff = 32^ feet, we have -=ij-^ hence on the surface of the earth a body which weighs 32^ pounds (nearly) is a unit of mass. It would be an exact unit if the acceleration were exactly 32 J feet. In order to illustrate this subject further, suppose that we make the unit of mass that of a standard jpound. Then equa- tion (19) becomes in which g^ is the value of g at the latitude of 45 degrees. This value resubstituted in the same equation gives g and these values in equation (18) give the final value of which is the same as (21). Again, if the unit of mass were the weight of one cubic foot of distilled water at the place where g^ = 32.1801 feet, at which place we would have W= 62.3791, and (19) would give _ 62.3791 ^' "32.1801' and this in the same equation gives 32.1801 W ^ ~ 62.3791 ' g ' and these values in (18) give „ Wd^s , . -^ = :t3 » as before. g dr' iO DENSITY :»,1 23' Deinsity is the mass of a unit ofvoVmae, li M= the mass of a body ; F= the volume; and J) = the density ; ttieii if the density is uniform, we have If the density is variable, let 8 = the density of any element, then .\M=fhdV (22) from which the mass may be determined when 8 is a known function of F. Examples. 1. In a prismatic bar, if the density increases uniformly from one end to the other, being zero at one end and 5 at the other, required the total mass. Let I = the length of the bar ; A = the area of the transverse section ; and X = the distance from the zero end ; then will 5 y = the density at a unit's distance from the zero end ; ! 5 -jx = the density at a distance x ; and dV=:zAdx, .\ M = A / —J— = - AL 2. In a circular disc of uniform thickness, if the density at a unit's distance from the centre is 2, and increases directly as the distance from the centre, required the mass when the radius is 10. {2L] APPLICATIONS OF EQUATION (21). 31 3. lu the preceding pn^blem suppose that the density in- creases as the square of the radius, required the mass. 4. In the preceding problem if the density is two pounds per cubic foot, required the weight of the disc. 6. If in a cone, the density diminishes as the cube of the distance from the apex, and is one at a distance one from the apex, required the mass of the cone. Having established a unit of density, we might properly say that mass is a certain number of densities, 24. Applications of Equation (21). [Obs. — If, for any cause, it is considered desirable to omit any of the matter of the following article, the author urges the student to at least establish the equations for the acceleration for each of the 31 examples here given. This part belongs purely to mechanics. The reduction of the equations belongs to mathematics. It would be a good exercise to establish the fundamental equa- tions for all these examples, before making any reductions. Such a course serves to impress the student with the distinction between mechanical and mathematical pnnciples.] 1st. If a hody whose weight is 50 jpounds is moved homon- tally hy a constant force of 10 pounds, required the velocity ao ^^m' quired at the end of 10 seconds ^^^^^^t^Z^^^ and the space passed over during ^^ ^ that time, th&re being no friction nor other external resistance, and the hody starting from rest. Here M ^ ^^ lbs., and . F= 10 lbs. ; hence (21) , gives d's F 193 30* Multiply by dt and integrate, and ds 193 ^^ = .= 3^^ + (^, = 0) The second integral is 29 ACCELERATING FORCES. raA.j and hence for ^ = 10 seconds, we have V = 64.33 + feet 8 = 321.66 + feet. 2d. Suppose the data to he the same as in the preceding example, and also tlmt the friction between the body and the plane is 6 pounds. Required the space passed over in 10 seconds. Here F= (10 - 5) pounds. # dJ's _F _ 193 *'d^^ M^ 60* Zd. Suppose that a body whose weight is 60 pounds is m^ved horizontally by a weight of 10 lbs,, which is attached to an i7iex- tensible, but perfectly flexible string which passes over a wheel and is attached at the other end to the body. Required the distance passed over in 10 seconds, if the string is without weighty and no resistance is offered by the wheel, plane^ or String, so lbs. -^ k JO Us. Fxo. 10. In this case gravity exerts a force of 10 pounds to move the mass, or i^= 10 lbs., and the mass moved is that of both bodies, or J/ = (50 + 10) ^ 32J-. ^_ i^^_193 df "M ~ 36 • The integration is performed as before. An^. s — 268.05 feet. \th. Find the tension of the styling in the preceding example. {34.1 MOVING MASSES. i8 The tension will equal that force which, if applied directly to the body, as in Ex. 1, will produce the same acceleration as in the preceding example. Let P = 10 pounds ; TT = 50 pounds ; T = tension ; P + TF = the mass in the former example ; and w — = the mass moved by the tension. 9 Hence, from Equation (21), -f=P\ and 9 Eliminate yj and we find WP r = W + P' /. r= 8.33 lbs. Wliat must be the value of P so that the tension will be a maximum or a minimum, P H- IF being constant? hth. In example 3, what must he the weight of P 80 thai the tension shall he (^) part of P f Ans. P = (71-1) W, Qth. If a hody whose weight is W falls freely in a vacuum hy the force of gravity, determine the formulas for the motion, llcre Mg = W and the moving force i^= W\ . TF^. _ -^. d^s ^^' 5? ^ ^- Tlie integrals of this equation will give Equations (16), when the initial space and velocity are zero. Let the student deduce them. 34 PROBLEks OF [24J 7 th, Suppose that the moving pressure {pull or pusJi) equals the weight of the body, required the velocity and space. Here Mg = W and i^= W, hence the circumstances of motion will be the same as in the preceding example. The forcjes of nature produce motion without apparent pressure, but this example shows that their effect is the same as that produced by a push or pull whose intensity equals the weight of the body, and hence both are measured by pounds y or their equivalent. Sth. If the force F is constant^ show that Ft = Mv ; also Fs = ^Mv^, and \Ff = Ms. If F is variable we havt Mv =fFdt. ^th. Suppose that a piston, d-evoid of friction, is driven hy a constant stea/m- pressure through a portion of the length of a cfylinder, at what point in the stroke must the pressure be instantly reversed Fig. 11. "^^^ ^0 that the full stroke shall equal the length of the cylinder, the cylinder heing horizontal f At the middle of the stroke. AVhatever velocity is g^n erated through one-half the stroke will be destroyed by the counter pressure during the other half. IQth. If the pressure upon the piston is 500 pounds^ weight of the piston bO pounds, and the friction of the piston in the cylinder 100 pounds, required the point in the stroke at which the pressure must be reversed that the stroke may be 12 inches. The uniform effective pressure for driving the piston is 600 — 100 = 400 lbs., and the uniform effective force for stopping the motion is 500 + 100 = 600 pounds. The velocity generated equals the velocity destroyed, and the velocity destroyed equals that which would be generated in the same space by a force equal to the resisting force ; hence if F= the effective moving force ; 8 = the space through which it acts ; V = the resultant velocity ; F' = the resisting force ; and s' = the space through which it acts ; P4.3 ACCELERATING FOROBS. then, from the expression in Example 8, we have Fs = i Jf^, and F's' = ^Mi^, .-. Fs = F'a', or, F '. F'\\8' \ 8, In the example, F= 400 lbs., and F'= 600 lbs. Let a? = the distance from the starting point to the point where the pressure must be reversed. Then 600 : 400 ::aj : 12 — «, .\ aj = 7^ inches. Wth, If in the jpreceding examjple thejpiston moves verticaUy ujp and down^ required the jpoint at which the pressure mt^t le instantly reversed so that the full stroke shall he 12 inches. The effective driving pressure upward will be 600 — 100 — ' 60 = 360 pounds, and the retarding force will be 600 4- 100 + 60 = 650 pounds, and during the down-stroke the driving force IS 600 -f 50 — 100 = 460 pounds, and the retarding force is 500 - 60 + 100 = 660 pounds. 12rA. A string passes over a wheel and has a weight P attached at one end, and W at the other. If there are no resistances from the string or wheel^ and the string is devoid of weight, required the resulting motion. Suppose W > P\ then . dh^ " dt^ Bj integrating, we find TT- P, and W+ P 9 F_ M W-P TF + P ^. and, V = 3 = i W-P W+P S^f W-P ^ irrp^^' 36 PROBLEMS OP [24] ISth, Required the tension of the atrmg in the j)7'eceding example. The tension equals the weight P, plus the force which will produce the acceleration when applied to raise P vertically. The mass multiplied by the acceleration is this moving force, or g' W-i-P hence the tension is P + W-P P = 2WP W+P ir+ p Similarly, it equals TF minus the accelerating force, or 2WP W+ P' w-Zrj:.w W+ P A complete solution of this class of problems involves the mass of the wheel and frictions, and will be considered here- after. 14^A. A string passes over a wheel and has a weight P attached to one end and on the other side of the wheel is a weight IF, which slides along the string, llequired the friction between the weight TF and the string^ so that the weight P will remain at rest. Also re^ quired the acceleration oftlie weight W. Fia. 13. hence, and, The friction = P ; Mg= TF; and,i^- TF-P; . oPs_ F^ ^ W-P "'' df M '^ W W-P g v = W giy «=4~-^p- g^^ t»l.| ACCELERATING FORCES. 27 Ibth. In the preceding example^ if W were an animal whose weight is less t/ian P, required the acceleration with which it must ascend, so tliat P will remain at rest. IQth. If the weight W descend along a rough rope with a given acceleration, required the acceleration with which tJi6 body P must ascend or descend 07i the opposite rope, so that the rope may remain at rest, no allowance being made for friction on the wheel, (The ascent must be due to climbing up on the cord, or be produced by an equivalent result.) VI th. A particle moves in a straight line under the action of a uniform acceleration, and describes spaces s and s' in t*** and t'^^ seconds respectively, determine the accelerating force and tJie velocity of projection. Let . Vo = the velocity of projection, and f=z the acceleration ; then t' --e s'{2t -■ 1) — 8(2^ - 1) and ^0=-^ ^^M) ' If - = -. , then Vq = 0. 8 2t — 1 IS fh. If a perfectly flexible and perfectly smooth rope w placed upon ajxin, find in wliat time it will run itself off , If it is perfectly balanced on the pin it will not move, unless it receive an initial velocity. If it be unbalanced, the weight of the unbalanced part will set it in motion. Suppose that it is balanced and let v^ — the initial velocity, 2^ = the leuirtli of the rope, w •= the weight of a unit of length, and t = the time. Talce the origin of coordinates at the end of the rope at the instant that motion begins. When one end has descended 8 feet, the other has ascended the same amount, and hence tho 2S PEOBLEMS OF 124.} 2 wi^ "• f unbalanced weight will be 2ws. 2wl -r g ; hence we have d^ _ F_ _2ws df~M~ Multiply by ds and integrate, and we ha^e The masB moved will be Integrating again, gives / 9 /7-o»+^ , if « = Z, 19^A. 7)^ ajpa/rticle moves towards a centre of force which ATTEAOTS dwectly as the distance from the force, determme the motion. Let fi = the absolute force ; that is, the acceleration at a unit's distance from the centre due to the force; and then s = the distance ; w=->^- A force is considered positive in whatever direction it acts, and the plus sign indicates that its direction of action is the same as that of the positive ordinate from the origin of coor- dinates, and the negative sign action in the opposite direction. If a be the initial value of s^ we have (see Ex. 5, Art. 17) : t = Li-i sm X--4-); PW.1 ACCELERATINa FORCES. 29 and the velocity at the centre of the force is fonnd by making « = 0, for which we have, V = a i^; 113 and t= - -^tr^rt, -^ir^n, g^"*^, etc., hence, the time is independent of the initial distance. It may be proved that within a homogeneous sphere the : attractive force varies directly as the distance from the centre. Hence, if the earth were sucli a sphere, and a body were per- mitted to pass freely through it, it would move with an accele- rated velocity from the surface to the centre, at which point the velocity would be a maximum, and it would move on with a retarded velocity and be brought to rest at the surface on the opposite side. It would then return to its original position, and thus move to and fro, like the oscillations of a pendulum. The acceleration due to gravity at the surface of the earth being g, and /• being the radius, the absolute force is and the time of passing from surface to surface on the equator would be t = .^- = 3.1416 \/~i^^}^ = 42m. 1.6 sec. The exact dimensions of the earth are unknown. The semi-polar axis of the earth is, as determined by Bessel 20,853,662ft. Airy 20,853,810 ft. Clarke 20,853,429 ft. The equatorial radius is not constant, on account of the elevations and depressions of the surface. There are some indications that the general form of the equator is an ellipse. Among the more recent determinations are those by Mr. Clarke, of England (1873), and his result given below is considered by him as the moat probable mean. The equatorial radius, is according to Bessel 20,923.596 ft. Airy 20,923,713 ft. Clarke 20,923.161ft. 30 PROBLEMS OF [24.] The di».termination of the force of gravity at any place is subject to small errors, and when it is computed for different places the result may differ from the actual value by a perceptible amount. The force of gravity at any particular place is assumed to be constant, but all we can assert is that if it is variable the most delicate observations have failed to detect it. But it is well known that the surface of the earth is constantly undergoing changes, being elevated in some places and depressed in others, and hence, assuming the law of gravitation to be exact and universal, we cannot escape the conclusion that the force of gravity at every place on its surface changes, and although the change is exceedingly slight, and the total change may extend over long periods of time, it may yet be possible, with apparatus vastly more delicate than that now used, to measure this change. It seems no more improbable than the solution of many problems already attained — such for instance, as determining the relative velocities of the earth and stars by means of the spectroscope. 20th. Suppose that a coiled spi'ing whose natural length u A B^ is co)jipressed to B C. If one end rests against an im- movable hody B, and the other against a hody at Cy which is perfectly free to 7nove horizontally, what wiU Fio. 14. he the time of m^ovement from C to A, and what will he the velocity at A ? It is found by experiment that the resistance of a spring to compression varies directly as the amount of compression, hence the action of the spring in pushing the body, will, in reference to the point A, be the same as an attractive force which varies directly as the distance, and hence it is similar to the preceding example. But if the spring is not attached to the particle the motion will not be periodic, but when the particle has reached the point A it will leave the spring and proceed with a uniform velocity. If the spring were destitute of mass, it would extend to A, and become instantly at rest, but because of the mass in it, the end will pass A and afterwards recoil and have aperiodic motion. If the body be attached to the spring, it will have a periodic motion, and the solution will be similar to the one in the Author's Resistance of Materials^ Article 19. Take the origin at A, s being counted to the left ; suppose that 6 pounds will compress the spring one inch, and let the total compression be « = 4 inches. Let W= the weight of the body = 10 pounds. [24.] ACCELERATING PORCES. 31 The force at the distance of one foot from the origin beiixg 60 pounds, the force at 8 feet will be 608 pounds. Hence, - ^ = - /^= - 60« ; g dt^ ^ = -193.; from which we find that t- ^ \/l93 and i; = 4.6 4- feet. 21st. Stippose that in the preceding jprohlem a body whose wetght is M' is at B, and another M " at C, both being perfectly free to move horizontally, required the time of movemeint that the distance between them shall be equal to AB ; and the resultant velocities of each. Take the origin at any convenient point, say in the line of the bodies and at a distance a?' to the left of Jff' , and let a;" be the abscissa of M" ; b the length of the spring after being compressed an amount «, and // the force in pounds which will compress it the first unit ; then the tension of the spring when the length is s will be ^(a + J — «) ; hence we have 8 • = x" - x\ ■jurud'^x" = f^{a + J — «X ^ df = -/^(a + J — > From the first, d^s d^ d}x" - d^ d^af substituting. dh df = ^ M'M" (^■*'^" -*); integrating, dl Jf'4 y>g- 32 REPULSIVE FORCES. 1^4.] Butv = Oiors = b; .-. C\ = - m^- -^^, ^; and integrating again gives ./ M' + M" . .s—h ,^, ^. and, making s = a + h, we have , , ./I J/' J/" which, as in the preceding example, is independent of the amount of compression of the spring. To find tLe relation between the absolute velocities, Let s' = the space passed over by Jf' , and s" = the space passed over by Jf ; then since the moving force is the same for both, we have integrating, gives 22d, Swppose that the force varies directly as the diaUmce from the centre of force and is repulsive. ;rhen d^s df = ^''^ in which v^ is the initial velocity. 23c?. Swppose that the force varies inversely as the square of the distance from the centre of the force and is attraotivk. [This is the law of universal gravitation, and is known as the law of the inverse sqwires. While it is rigidly true, so far as we know, for e^eiy [24. \~ ATTRACTIVE FORCES. 38 particle of matter acting upon any other particle, it is not rigidlj' true for finite bodies acting npon other bodies at a finite distance, except for homoge- neous spheres^ or spheres composed of homogeneous shells. The earth bein^ neither homogeneous nor a sphere, it will not be exactly true that it attracts external bodies with a force which varies inversely as the square of the distance from the centre, but the deviations from the law for bodies at great distances from the earth will not be perceptible. We assume that the law applies to all bodies above the surface of the earth, the centre of the force being at the centre of the earth.] Let the problem be applied to the attraction of the earth, and r = the radius of the earth ; g = the force of gravity at the surface ; fi = the absolute force ; and 8 = the distance from the centre ; then and f* = g^ Multiply by ds and integrate ; observing that for « = a, v = and we have -f ^ as — s^ '2ju \i , _ — sds using the negative sign, because t and 8 are inverse functions of each other. The second member may be put in a convenient form for integration by adding and subtracting ^ a to the numerator and arranging the terms. This gives ^a — 8 — ^a (08 - «2)i ds a — '2,8 J ads ds — 2(as-^)i 2{a8-8^ ' 84 ATTRACTIVE FORCES. rS4i|> the integral of which is (as — s^)i — ia versin""* — + O. ^ ^ a But when s = a, t = .'. 0= ^a^ ; I From the circle we have tt — versiu"' — = n — coerwl — _ \ =? coe-.( -(l-r))=--(V^)- From trigonometry we have 2 cos^ y — 1 = cos 2 y. Let 2 y— cos ~' [ 1 1, then cos 2y = -f — 1 ; . \ cos^^ y = —i and y = cos-' 4 /£ ; and a a y a 2y = 2 cos-' /</ — ; or TT — versin-'— J. From {a) it appears that for 5 = 0, -y = 00 ; hence the velocity at the centre will be infinite when the body falls from a finite distance. If5 = a = oo,'y = 0. If a body falls freely from an infinite distance to the earth, we have in equation (a) ^ = 00 ; and s = /• = tlie radius of the earth ; •■-(4r')'. for the velocity at the surface. I^"t ^ —9\ :. V = {2gr)h If ^ = 32J feet and r = 3962 miles, we have /64J X 3962U . ^. ., , Hence the maximum velocity with which a body can reacb the earth is les& than seven miles per second. fSll ATTBACTIVE FORCES. 35>^ 24:M. Sujypose that the force is attraotivb and vanes in versely as the n*^ jpower of' the distance. Then ^ =ljt- (1 L V and integrating, gives According to tlie tests of integrahility this may be integrated when 6 3 1.35 n. = ....-=,-, g , — 1, —, or - ... . etc., 3 2 1^^3 orn= ....-,-, -,0,2, or- etc 25 ^A. Let tJie force vary inversely as the square root of the distance and he attractive. (This is one of the special of the preceding example.) We have or, 2 y^dt ~ ^ (a» - «♦)* The negative mgn is taken because t and s ore inyene fonctions of each oibei. Add and aubtract — ■;^ — ; — - and we have 3 yi yak-sk /- F - Vr 2v/a 2v/7 1 86 ATTRACTIVE FORCES. [24.] L S/y/T/y/ ai _ *J 3/y/Tv' tt* -«*J 26^A. Sv^ose that the force is atteaotive cmd varies ify versely as the distance. Hence 'de" 7' , d^ o 1 « in which s — a for t; = 0. Hence the time from = a to # = 0, is Let I log — 1 = y ; then for « = a, y = and for « = 0, y = ». Squaring and passing to exponentials, we have log — = t^ . •.— = 6^ , or « = a e^ ; .'. d8 = — ae~^ .2y dy. This is called a gamma-function^ and a method of integrating it is aa follows : — Since functions of the same form integrated between the same limits az» independent of the variables and have the same value, therefore f34.J ATTRACTIVE FORCES. 87 and Also the left hand member will be of the same value if the sign of integr»- Hon be placed oyer the whole of it, since the actual integration will be performed in the same order ; hence 2 . . -^dydt [/•^'•]\r/- */<? c/<? •00 2 «.-''(!+»'• <«*.5 o ^ o hi which y = ^w ; . ♦. dy = td\u Integrating in reference to *, we have _^-«2 (1 + w2) 2 (1 + w2) du tegral of this is \ tan~' «, which is zero for i^ = 0, and i tt f or i* = oo ; which for < = 00 becomes zero, and for t = becomes r- , and the in- 2(1 4- tr) ../.-•*= V- (See also Wee. Celeste, p. 151 [1534 O].) Or we may proceed as follows : — L&t e-t^ = X .: dt=-{-\osxT^^i . •. / e-^^ dt= / - i (- log x)-^ dx. Jo J\ Let x — aif and consider a less than unity ; then log a will be negative, .'. dx=ay 2y dy{— log a) ; and log« = /(-loga); 88 ATTRACTIVE FORCES. [34.1 which substituted above gives Jq ert^dt =J - i ( - log x)-idx =: ( - log a)^ i a^ dy . IMviding by ( — log ay and multiplying both sides by — ^ da, we have J -\ (- log aT^da J -i (- log xT^dK =j J -i a^% da. Integrating the second member first in regard to a, gives /8 +1 -I y2-4-i which between the limits of and 1 gives i ; the integral of which is t + p^ I tan-iy which between the limits of oo and gives irr. .-. y - i (- log a)"* da J -i (- log x)-^d» •*• / e~* dt = iVn: (See Mie. CHeste, Vol iv. p. 487, Nos. [8819] to .[8331]. Chauvenet's /Spherical Astronomy^ Vol i. p. 152. TodhnnteT' s Bitegral Caleidus. Price's Infinitesimal Calcultis.) Sometimes the integration of an exponential quantity becomes apparent by first differentiating a similar one. Thus, to integrate t e~^ dt, first differ- entiate e-*^' We have d er^z= e-^d {- t^) = e-^^ i-2tdt) = - Ue-^ dt ; . •. / de-^ = -2 I te-^ dt. But the first member is the integral of the differential, and hence is the .2 2 quantity itself, or €~'' , and hence the required integral is -^ e"'' . 27^A. Suppose that two bodies have their centres at A and A' respectively, and- ^^ C I A _^' ATYRkCT a particle "at Fio. 15. j> with forces which vary as the distances from A and A', /J»l] ATTBACTIVE FOECEa 39 Let C be midway between A and A' ; AC=^ OA'=a; Ch — 8=^ any variable distance ; and let fi = /<' = the absolute forces of the bodies A and A respectively. Then ^ = ,. (a -») -M (a + «) = - 2/t»; and integrating again gives 8-=z c cos ^ \/2^. 28^A. Sujypose that ajparticle isjprojected with a velocity u into a medium which resists as the square of the velocity / determine the circumstances of motion. Take the origin at the point of projection, and the axis 8 to coincide with the path of the body. Let fi = the absolute resistance — or the resistance of the medium when the velocity is unity ; then f« I;!-) = the resistance for any velocity ; d^8 _ /dsy or, _^ = -^. dt da And integrating between the initial limits, g = for -— = i*, and the general limits, we have "^^ RESISTING FOBCES. f»4.> or, log — = — fi« ; u ds -us or, vdt = ^ ds. Integrating again, observing that t — for « = C, we have The velocity becomes zero only when 5 = oo . 29^A. ^ A^^-yy hody falls in the air hy the force of gravity^ the resistance of the air varying as the square of the velocity / determine the motion. Take the origin at the starting point, and Let K = the resistance of the body for a unit of velocity ; s = the distance from the initial point, positive down- wards ; t = the time of falling through distance 8 ; ds then ;t7 = for t and s = ; -rr = V for t = t and 5 = 5; and dt k{ — \= the resistance of the air at any point, and acts upwards ; and g = the accelerating force downward ; henco, the resultant acceleration is the difference of the two, or d^s (dsY , . \dt) [94.] FALLING BODIES. il ... '© * U/ Separating thia into two partial fractions, and integrating^ gitw Passing to exponentials giyes (») which gives the Telocity in terms of the time. To find it in terms of tho space, multiply equation (a) by da and put it under the form j = 2Kdt, Proceeding as before, observing the proper limits, we find 'd8\9 9 = - log - -Q ••J— t/f(<--"-) (<5) =v1. If« = oo,r = 4/-, and hence the velocity tends towards a constant. From equatiop (ft), multiplying the terms of the fraction by «— ^ ("5')*. and observing that the numerator becomes the differential of the denominater integrating, and passing to exponentials, we have, which gives the space in terms of the time. A neat solution of equation (a) may be found by Lagrange's method of Variation of Parameters. ^2 PALLING BODIES. [34.1 SOth, Suppose that the hody is projected wpward in the ai/r^ having the same coefficient of resistance as in the jyreceding example. Take the origin at the point of propulsion, u being the initial velocity ; then dJ^s d ds __ (dsV de^^'dt di~ " ^ '^ ""Xdt) '' hence, Kdt = KXdt) Solving this equation for — ; we have, _ ^ _ /pU uV K-\fg tan t V ng ~dt-\K/ Vi+uV^ t^ntV^g' ^^ Substitute sin t V Kg -i- cos t V Kg tor tan tVKg and the numerator bo- <Oomes the differential of the denominator, and observing that t = for s = 0, we have 1 , w Vk sin t y/ Kg + ^ g cos t "sf kq « = — log 7= ; which gives the space in terms of the time. Multiply equation {e) by ds and it may be put under the fonn 2Kd8 = K ^ \dt) lat^^ting, observing that u is the initial velocity, and fds\^ 3 + ^ {d) ^'^=-'og -— ^ g + K"'^ * 124.] IN A RESISTING MEDIUM. 43 At the highest point v = 0, which in (/) and (g) gives Substitute this value of s in equation (c) of the preceding example, and we have = / ''(- 1 N -{' 9 ' 9 u; g + >^U2 which gives the velocity in descending to the pohit from which it started ; and as it is less than u, the velocity of return will be less than that with which it was thrown upward. This is because the resistance of the air is against the velocity during the entire movement, both upwards and down- wards. The same value of « (Eq. (i) ) substituted in {d) of the preceding example gives the time of descent, / = ^-4= log i^^^iLVJ; which differs from the time of the ascent, as given by {h) above. ^Ist. Suppose that the force is attractive and varies inversely as the cube of the distance, and that the medium resists as the square of the velocity, and as the square of the density, the density varying inversely as the distance from the origin. Let H = the coefficient of resistance, being the resistance for a unit of density of the medium and a unit of velocity ^ WORK. f25.> then -^ \Zj7) = *h® resistance at any point. dh u x (dsY df «8 Multiply by Ms, and we have This is a linear differential equation of which the integrating factor is 6 . The initial values are t = 0, and 8 = atoTV = u; -■©-"■■ =&n 2k — « — 2k — a e s — - e a a which gives the velocity in terms of the space. The final integral cannot be found. 25. Work and Yis Yiva {or living force.) — Resuming equation (21), and multiplying both members by ds, we have Fds = M^ds, df integrating between the limits, v — v^ for 5 = 0; and v = t; for 5 = 5, we have jFds = Wi^ - %^)' (23) If Vq — 0,we have fFds = W^' (24) The expression i3fiy^ is called the vis viva (or living force) of a body whose mass is M and velocity v. Its physical im- portance is determined from tlie first member of the equation, which is called the work done by a force T^^in the space s. Hence the vis viva equals the work done hy the movivbg force. Work, mechanically, is overcoming resistance. It requires a certain amount of work to raise one pound one foot, and twice {25.J WORK. 45 that amount to raise two pounds one foot, or one pound two feet. Similarly, if it requires 100 pounds to move a load on a horizontal plane, a certain amount of work will be accomplished in moving it one foot, twice that amount in moving it two feet, and so on. Hence, generally, if F= B. constant force which overcomes a constant resist- ance, and s = the space over which Tracts projected on the action- line of the force, then Workz=z U=F8; (25) and similarly, if /^= a va/riahle force, then Work= U=XFds; (26) and if i^ is a function of 8 we have U=fFd8. The UNIT of work is one pound raised vertically one foot. The total work, according to equation (25), is independent of the time, since the space may be accomplished in a longer or shorter time. But implicitly it is a function of the time and velocity. If the work be done at a uniform rate, we have and If f = 1, we have 8 = Vt, Fs = Fvt, Fv, (28) which is called the Dynamic Effect^ or Mechanical Power, Mechaihcal Power is the rate of doing work. It is meas- ured by the amount of work done, or which the agent is capa- ble of doing, in a unit of time when working uniformly. The unit most commonly employed is called the horae-powevy which equals 33,000 pounds raised one foot per minute. Evexy moving body on the surf ace of the earth does work, for it oyercomet a resistance, whether it be friction or resistance of the air, or some other reibtance. The same is true of every body in the universe, unless it movea 46 WORK. [25. J In a non-resisting medium. * Animals work not only as beasts of burden, but in th6ir sports and efforts to maintain life ; water as it courses the stream wears its banks or the bed, or turns machinery ; wind fills the sail and drives the vessel, or turns the windmill, or in the fury of the tornado levels the forest, and often destroys the works of man. The raising of water into the air by means of evaporation ; the wearing down of hills and mountains by the operations of nature; the destruction which follows the lightning-stroke, etc., are examples of work. Work may be useful or prejudicial. That work is useful which is directly instrumental in producing useful effects, and prejudicial when it wears the machinery which produces it Thus in drawing a train of cars, the useful work is performed in moving the train, but the prejudicial work is overcoming the friction of the axles, the friction on the track, the resistance of the air, the resistance of gravity on up grades, etc. It is not always possible to draw a practical line between the useful and prejudicial works, but the sum of the two always equals the total work done, and hence for economy the latter should be reduced as much as possible. In order to determine practically the work done, the inten- Bity of the force and the space over which it acts mu&t be measured simultaneously. Some form of spring balance ia commonly used to measure the force, and when thus employed is called a Dynamometer. It is placed between the moving force and the resistance, and the reading may be observed, or autographically registered by means of suitable mechanism. The corresponding space may also be measured directly, or secured automatically. There are many devices for securing these ends, and not a few make both records automatically and Bimultancously. If the force is not a continuous function of the space, equa- tion (2G) must be used. The result may be shown graphically by laying off on the abscissa, AB, the distances ac, ce^ etc., proportional to the spaces, and erecting oj'dinates ab^ cd^ ef^ etc., proportional to the corresponding forces, and joining their upper ends by a broken line, or, what is better, by a line which * All space is filled with something, since light is transmitted from all directions. But is it not possible that there may be a something throngh which bodies may mo^'e without resistance ? [25.1 WOKK. 47 is slightly curved, the amount and direction of curvature being indicated by the broken line previously constructed ; and the area thus inclosed will h^ A Fio. 16. represent the work. ^ ji The area will be mv- A I I en by the formula Simpson's rule for determining the area is: — Divide the abscissa AB into an even number of equal jpartB^ erect ordlnates at the points of division^ and number them in the order of the natural numbers. Add together four times ike even ordinates^ twice tJie odd ordinates and tlie extreme ordlnates, and mxdtiply the sum by one third of the distance between any two consecutive ordinates. If yi, ^2? ysj 6tc., are the successive ordinates, and I the dis- tance between any two consecutive ones, the rule is expressed algebraically as follows : — Area = il {y, + 4y, + 2y., -h 4:?j, + y„). (29) If the applied pressure, Jp] is exerted against a body which is perfectly free to move, generating a velocity v, then the work which has been expended is, equation (24), ^Mtr^. This is called stored work, and the amount of work whicli will be done by the moving body in being brought to rest will be the same amount. If the body is not perfectly free the quantity ^Mv^ is the quantity of work which has been expended by so much of the a])plied force as exceeds that which is necessary in overcoming the frictional resistance. Thus a locomotive starts a train from rest, and when the velocity is small the power exerted by the locomotive may exceed considerably the resist- ances of friction, air, etc., and produce an increasing velocity, until the resistances equal constantly the tractive force of the locomotive, after which the velocity will be uniform. The work done by the locomotive in producing the velocity v in excess of that done in overcoming the resistances will be iMv^y in which Jf is the mass of the train, including the locomotive. 4:8 EXAMPLES. [85.J We see that double the velocity produces four times the work. This is because twice the force produces twice the velocity, and hence the body will pass over twice the space in the same time, so that in producing double the velocity we ha\ e 2i^25 = 4jF5, and similarly for other velocities. [We have no single word to express the unit of living force. If a unit of mass moving with a velocity of one foot per second be the unit of living force, and be called a Dynam^ then would the living force for any velocity and mass be a certain number of DynamsJ] Since work is not force, but the effect of a force exerted through a certain space, independently of the time, we call it, for the sake of brevity, s^aoe-effeGt. Vis viva, or living force, is not force, but it equals the work stored in a moving mass. It equals the space-effect. [The expression Mi? was called the vis viva in the first edi- tion of this work, and is still so defined by many writera ; but there appears to be a growing tendency towards the general adoption of the definition given in the text. It is immaterial which is used, provided it is always used in the same sense.] Examples. 1. A body whose weight is 10 pounds is moving with a velocity of 25 feet per second ; required the amount of work which will be done in bringing it to rest. Arts, 97.2 foot-pounds. 2. A body falls by the force of gravity through a height of h feet ; required the work stored in it. Let W = the weight of the body, M = the mass of the body, g = acceleration due to gra ;ity, and V = the final velocity, then t? = 2gh, B.nd Mg = W; W -96. J ENERGY. 49 3. A body whose weight is 100 pounds is moving on a hori- zontal plane with a velocity of 15 feet per second ; how far will it go before it is brought to rest, if the friction is con- stantly 10 lbs ? Ans. = 34.9 + ft. 4. A hammer whose weight is 2000 pounds has a velocity of 20 feet per second ; how far will it drive a pile if the constant resistance is 10,000 pounds, supposing that the whole vis mva is expended in driving the pile ? 5. If a train of care whose weight is 100,000 pounds is moving with a velocity of 40 miles per hour, liuw far will it move before it is brought to rest by the force of friction, the friction being 8 pounds per ton, or -^fj^ of the total weight ? 6. If a train of cars weighs 300 tons, and the frictional resistance to its movement is 8 pounds per ton ; required the hoi-se-power which is necessary to overcome this resistance at the rate of 40 miles per hour. Am. 256. 7. If the area of a steam piston is 75 square inches, and the steam pressure is 60 pounds per square inch, and the velocity of the piston is 200 feet per minute, required the horse-power developed by the steam. 8. If a stream of water passes over a dam and falls through a vertical height of 16 feet, and the transverse section of the stream at the foot of the fall is one square foot, required the horse-power that is constantly developed. liCt g =32| feet, and the weight of a cubic foot of water. 62i lbs. Ans. 58.2 4 9. A steam hammer falls vertically through a height of 8 feet under the action of its own weight and a steam pressure of 1000 pounds. If the weight of the hammer is 500 pounds, required the amount of work which it can do at the end of the fall. 26- Energy is the capacity/ of an agent for doiTig work. The energv of a moving body is called actual or Kinetic energy^ and is ex})ressed by \M'i^. But bodies not in motion mav have 4 50 MOMENTUMo f27.1 a capacity for work when the restraining forces are removed. Thus a spring under strain, water stored in a mill-dam, steam in a boiler, bodies supported at an elevation, etc., are exam- ples of stored work which is latent. This is called Potential energy. A moving body may possess potential energy entirely distinct from the actual. Thus, a locomotive boiler containing steam, may be moved on a track, and the kinetic energy would be expressed by \Mv''^ in which M is the mass of the boiler, but the potential energy would be the amount of work which the steam is capable of doing when used to run machinery, or is otherwise employed. Tliese principles have been generalized into a law called the Conservation of energy^ which implies that the total energy, including both Kinetic and Potential, in the universe remains constant. It is made the fundamental theorem of modern physical science. The energy stored in a moving body is not changed by changing the direction of its path, provided the velocity is not changed ; for its energy will be constantly expressed by | J/i>'. Such a change may be secured by a force acting continually normal to the path of the moving body ; and hence we say that a force ■which ael.9 co/Ui/iually peipendicalar to the p)(ith of a ifwviny hody does no work upon, t/te Ixxly. Thus, if a body is secured to a ])oint. by a coi-d so that it is compelled to move in the circuniferent^e of a cii'cle ; the tension of the 8tring does no work, and the vis viva is not affected by the body being constantly detiecited from a rectilinear path. MOMENTUM. 27- Itcsuming again equation (21), multiplying by dt^ and integrating gives, /^^' = ^^/w = ^5-^^- (30) The expression Mv is called nuynientuin^ and by comparhig it with the fii-st member of the ecpiation we see that it is the effect of the force F acting during the time ^, and is indepen- dent of the s])ace. For the sake of brevity we may call the momentum a time-effect. [27.1 MOMENTUM. 51 If the body lias an initial velocity we have Fdt = M(v-Vo); (31) / to which is the momentum gained or lost in paesing from a velocity Vq to v, Momentmn is sometimes called quantity of motion^ on account of its analogy to some other quantities. Thus the intensity of heat depends upon temperature, and is measured in degrees; but the quantity of heat depends upon the volume of the body containing the heat and its intensity. The iiften- sity of light may be uniform over a given surface, and will be measured by the light on a unit of surface; but the quantity is the product of the area nniltiplied by the intensity. The uitensity of gravity is measured by the acceleration which is produced in a falling body, and is independent of the mass of the body; but the quantity of gravity (or total force) is the product of the mass by the intensity (or Mg). Similarly with momentum. The velocity represents the intensity of the motion, and is independent of the mass of the body ; but thtP quantity of motion is the i)r()duct of the mass Multiplied by the velocity. Differentiating (30) and reducing, gives which is the same as (18), and in which -7 is a velocity-incre- ment ; Jience the rate of change of the momentum per unit of time is a measure of the force which is acting on (he body. If the force I*^ is constant we have from (30), M = 21 V ; and for another force F" acting during the same time F't^M'v'\ .-. F',F'\\ Mo: M'v'-, hence, tiie forces are directly as the momenta produced by them respectively. 52 IMPACT. pa.] If the forces are variable, let CFdt =Q = Mv, and f F'dt = Q' ^ i/V; then Q\Q' wMv. M'v' ; hence the time-effects are directly as the momenta impressed. We thus have several distinct quantities growing out of equation (21) of which the English units are as follows : — ^e unit of force, j^, is 1 ft. The unit of work or space effect is IBt) xlft. The unit of vis viva is 1 Tb of mass x 1^ ft. x 1 sec. The unit of momentum , 1 tb of mass x 1 ft. x 1 sec. IMPULSE. 28, An impulse is the effect of a blow. When one body strikes another, an impact is said to take place, and certain effects are produced upon the bodies. These effects are pro- duced in an exceedingly short time, and for this reason they are sometimes called instantaneous forces ; which, being strictly defined, means a force which produces its effect in- stantlyy requiring no time for its action ; but no such force exists in nature. The law of action during impact is not gen- erally known, but it must be some function of the time. Resuming equation (31), we liave j Fdt = M{v - vo) ; to which is true, whatever be the relation between the force i^^and the time t. If the initial velocity of the body be zero, we have Vo = 0, and / Fdt = MV= Q, -^to We see from the above equation that as t diminishes F must increase to produce the same effect. We see that in this [28.] IMPULSE. 63 case the first member is the time-effect of an impulse, and the second member measures its effect in producing a change of velocity. Calling this value Q^ we have Q = M{:p- v^ = MV. (31a) Hence, the mecbsare of an invpuUe in jproducing a change of velocity of a hody is the increased {or decreased) momentum jfyi'odticed in the hody. This is the same as when the force and time are finite. If the force were strictly instantaneous the velocity would be changed from Vq to v without moving the body, since it would have no time in which to move it. Similarly from equation (23) we have r Fds = iM{v'-Vo'); in which for an impulse JF* will be indefinitely large; and hence the work done hy an impulse is measured in the sam$ v)ay as f 07' finite forces. All the effects therefore of an impulse are measured in the same way as the total effects produced by a finite force. Iii regard to forces^ we investigate their laws of action ; or having those laws and the initial condition of the body we may determine the velocity, energy, or position of the body at any instant of time or at any point in space, and hence we may determine final results ; but in regard to impulses we deter- mine only certain final results without assuming to know any- thing of the laws of action of the forces, or of the time or space occupied in producing the effect. The terms ^''Impulsive force, " and ''^Instantaneous force^' are frequently used to denote the effect of an Impact ; but since the effect is not a force, they are ambiguous, and the term Impulse appears to be more appropriate. An incessant force may be considered as the action of an infinite number of infinitesimal impulses in a finite time. The question is sometimes asked, " What is the force of a 54 EXAMPLES. [28. J blow of a hammer ? " If by the force is meant the pressure in pounds between the face of the hammer and the object struck, it cannot be determined unless the law of resistance to compression between the bodies is known during the con- tact of tlie bodies. But this law is generally unknown. The pressure begins with nothing at the instant of contact and increases very rapidly up to the instant of greatest compression, after which the pressure diminishes. The pressure involves the elasticity of both bodies ; the rapidity with which the force is transmitted from one particle to another ; the amount of the distortion ; the pliability of the bodies ; the duration of the impact ; and some of these depend upon the degree of fixed- ness of the body struck ; and several other minor conditions; and hence we consider it impossible to tell exactly what the force is- EXAMPLES. 1. Two bodies whose weights arc IF and W^ are placed very near each other, and an explosive is discharged between them ; required the relative velocities after the discharge. 2. A man stands upon a rough board which is on a perfectly smooth plane, and jumps off from the board ; required the relative velocities of the man and board. [Obs. The common centre of gravity of the man and board will remain the same after they separate that it was before. After separatiig they would move on forever if they did not meet with any obstacle to prevent their motion.] 3. A man whose weight is 150 pounds walks from one end of a rough board to the other, which is twelve feet long, and free to slide on a perfectly suiooth plane ; if the board weighs 50 pounds, required the distance travelled by the man in space. 4. In example 3 of article 24, suppose that the weight 10 pounds is permitted to fall freely through a height A, when it produces an impulse on the body (50 pounds) through tlie intei mediate inextensible string ; required the initial velo<;ity of the body. '328. J EXAMPLES. 55 Let Vq = \/2yA = the velocity of the weight just before the impulse ; and V = the velocity iminediately afterward, wliich will be the comiiion velocity of the body and weight ; then 50 10 Q = -jV = - (vo-v); The subsequent motitjn may be found by equation (21), observing that the initial velocity is v. The tension on the string will be infinite if it is inextensible, but pi-actically it will be Unite, for it will be more or less elastic. [Some writers have used the expression impulsive tension of the strinf instead of nunnentum.] 6, If a shell is moving in a straight line, in vacuo, with a velocity v, and bui-sts, dividing into two parts, one part moving directly in advance with double the velocity of the body ; what must be the ratio of the weights of the two parts so that the other part will be at rest after the body bursts? 6. Explain how a person sitting in a chair may move across a roirnn by a series of jerks without touching the floor. (Can he advance if the floor is perfectly smooth ?) 7. A person is placed on a perfectly smooth plane, show liow he can get off if he cannot reach the edge of the plane. The same impulse ap])lied to a small body will impart a greater amount of energy than if applied to a large one. Thus, in the discharge of a gun, the impulse imparted to the gun ecjuals that imparted to the ball, but the work, or destruc- tive effect, of the gun is small compared with that of the ball. The tiifie of the action of the exj)losive is the same upon both bcnlies, but the e])ace moved over by the gun will be small coni]»Mred with that of the ball during that time. The j>roduct Mv^ being the same for both, as M decreases « incieases, but the work varies as the square of the velocity. 56 IMPACT. [394 DIBECT CENTRAL IMPACT. 29. If two bodies impinge upon one another, so that the line of motion before impact passes through the centre of the bodies, it is said to be central ; and if at the same time the common tangent at the point of contact is perpendicular to the line of motion, it is said to be direct and central. If their common tangent is perpendicular to the line of motion, but if the latter does not pass through the centre of the body impinged upon^ it is called eccentric impact. In this place, we consider only the simplest case ; that in which the impact is direct and centrals When two bodies impinge directly against one another, whether moving in the same or opposite directions, they mutually dis- place the particles in the vicinity ^ P of the point of contact, producing compression which goes on increas- % ing until it becomes a maximum^ at which instant they have a com- FiG. 17. mon velocity. A complete analysis of the motion during contact in- volves a knowledge of the motion of all the particles of the mass, and would make an exceedingly complicated prc^blem, bnt the motion at the instant of maximum compression may be easily foimd if we assume that the compression is instantly dis- tributed throughout the mass. Let Ml and M^ be the respective masses of the bodies ; Vi and V2 the respective velocities before impact, both positive and Vx>V2 ; Vi and V2 the respective velocities at the instant of maximum compression, v'^ > '^2? ^-^d Qi and Q^ the momenta gained respectively by the bodies during compression. Then from (31) which will be the momentum gained by M2 on account of the action of Mi. Similarly Qi = Mi{v^ -vi\ which, being negative, will be the momentum lost by Mi on account of the action of M<,. 180.] IMPACT. But at the instant of greatest compression < = -^i ; and, because they are in mutual contact during the same time, their time-effects are equal, but in opposite directions, .■■Qx=-Q^- Combining these four equations, we find by eUmination ^' = -MV^^M, ' = ^^ ' (^") which velocity remains constant for perfectly non-elastic bodies after impact, since such bodies have no power of restitution and will move on with a common velocity. DIRECT CENTRAL IMPACT OF ELASTIC BODIES. 30. Elastic bodies are such as regain a part or all of their distortion when the distorting force is removed. If they regain their original form they are called ^er/^c^/y elastic^ but if only a pai't, they are called imperfectly elaatic. After the impact has produced a maximum compression, the elastic force of the bodies causes them to separate, but all the effect which the force of restitution can produce upon the movement of the bodies, evidently takes place while they are in contact. If they are perfectly elastic and do not fully regain their form at the instant of se]»aration, they will continue to regain their form after separation, but the latter effect we do not consider in this place. The ratio between the forces of compression and those of restitution has often been called the modulus of elasticity ^ but as some ambiguity results from this definition, we will call it the modulus of restitution. At every point of the restitution there is assumed to be a constant ratio between the force due to compression and that to restitution. But it is unnecessary for present purposes to trace these effects, for by equation (31) we may determine the result when the bodies finally separate from each other. 58 IMPACT. [51. Let «| = the ratio of the force of compression to that of restitution of one body, which is called the modulus of restitution, e^ = the corresponding value for the other; Vi = the velocity of Mi at the instant when they separ- ate from each other ; and V2 = the corresponding velocity for M2. Then from equation (31) eiQi = 3fi{V,-Vi'); (34) e,Q,= M,{V,-n/). (35) As before Qi= — Qz and we will also assume that ^ = ^^ = ^. These combined with (32) and (33) give 81. Discussion of Equations (36) and (37). 1". If the bodies are perfectly non-elastic, « = 0. which is the same as (33). 2°. If the restitution is perfect « = 1. ■•••^' = '''-3rSi7, ('''-''») 5 (39) 2 J/" F, = «, + jnf^^'jf^ («'. - «>»). (40) 181.1 IMPACT. 59 From (38) we have M and, V,-v,= ^^^ K-t^ Similarly from (39) and (40) hence, the vcloc.it}' lost In- one body and gained by the other ia twice as much wlien tlie bodies are i)erfcctly elastic as when they are i»crl"cctly n on -clastic. tS°. If Ml = J/2, then for i)crfcct restitution we have 17 2J/, . ^^ = ^'^ ^ -jttttt; (^^-^•^) = ^^5 that is, they will intcrclianpje velocities. 4°. If Ml inipin«^cs against a fixed body, we have J/^ = oo, and V2 = 0. .*. Vi = -evi, This furnishes a convenient mode of determininp^ e. For if a body falls from a height A upon a Hxed horizontal plane, it will rebound to a liciglit A, ; /. hi = ^A, or d = W -?^. Also if tf = 1 or the velocity after impact will be the same as before, but m an opposite direction. 00 EXAMPLES. [32.1 Also if € = 0, Fi = ; or the velocity will be destroyed. 5°. If iX| = we have V.^^^^v, > (41) Examples. (1.) A mass Mi with a velocity of 10, impinges on M^ moving in an opposite direction, moving with a velocity 4 and has its velocity reduced to 5 ; required the relative magnitudes of Mi and Jfa. (2.) Two inelastic bodies, weighing 8 and 5 pounds respectively, move in the same direction with velocities 7 and 3 ; required the common velocity after impact, and the velocity lost and gained by each. (3.) If Ml weighs 12 pounds and moves with a velocity of 15, and is im- pinged upon by a body J/2 weighing 16 pounds, producing a common velocity of 30, required the velocity of M2 before impact if it moves in the same or opposite direction. (4.) If 5i¥"i = QM2 , 6«i = — 5v2 , ®2 = 7, and 6 = | ; required the velocity of each after impact. (5.) If i/i = 2if2 , Vi = |«i, and ?)2 = ; required e. (6.) If Vi is 26, M2 is moving in an opposite direction with a velocity of 16 ; Ml = 2M'i ,6=1; required the distance between them 5^ seconds after impact. (7.) Two bodies are perfectly elastic and move in opposite directions ; the weight of Ml is twice ifa, but ^2 = 2«i ; required the velocities after impact. (8.) There is a row of perfectly elastic bodies in geometrical progression whose common ratio is 3, the first impinges on the second, the second on the third and so on ; the last moves off with -^ the velocity of the first. What is the namber of bodies ? Ans, 7. LOSS OF VIS VIVA EST THE IMPACT OF BODIES. 32, Before impact the vis viva of both bodies was and after impact [33.] RELATIONS OF FORCE, MOMENTUM, AND WORK. 6J which by means of (36) and (37) becomes M, F,> -f M, V,' = M,v} + Mio{ - (l^-^^(t;,-t,,)«. (42) For perfectly elastic bodies e — \ and the last term disap- pears ; hence in the impact of perfectly elastic bodies no vis viva is lost. If the bodies are imperfectly elastic e is less than 1, and smce " ^-y^ _ ^^^ is always positive, it follows that in the impact of iAnperfectly elastic hodies vis viva is always lost, and the greatest loss is suffered when the bodies are perfectly noTirelastic, If 6 = 0, (42) becomes M,(v,» - V^) + Jf, W - V^) = ,^=^^ (V, - v,f ; (43) in which each member is the total loss by both bodies. It is also the loss up to the instant of greatest compression when the bodies are elastic. If ii/i is very large compared with M^ we have from (38) Fi = v^ nearly, = Fj , and (43) becomes M,v,^ -^ M,V,^ = M,{v,-y,f, the second member of which is frequently used in hydraulics for finding the vis viva lost by a sudden change of velocity. These investigations show the great utility of springs in vehicles and machines which are subjected to impact. RELATIONS OF FORCE, MOMENTUM, WORK, AND VIS VIVA. 38. We mat NOW determine the exact office in the same problem of the quantities \—force^ mofnentum, work, and via viva. Suppose that a force, whether variable or constant, impels a body, it will in a time t generate in the mass M a certain velocity v. T\i\9> force may at any instant of its action be measured by a certain number of pounds or its equivalent. 62 IMPULSE. STATICS. [34,85.1 Suppose that this mass impinges upon another body, wliich may be at rest or in motion. In order to determine the effect upon their velocities we use the principle of 7nomentum^ as has been shown. But the bodies are compressed during impact and hen^-e work is done. The amount of work which they are capable of doing is equal to the sum of their vis vwa /.and if they are brought to rest all this work is expended in compress- ing them. If the velocity of a body after impact is less thar. tliat before, it has done an amount of work represented by ■^3/{v^ — F^), and similarly if the other body has its velocity increased kinetic energy is imparted to it. The distortions of hodies represent a certain amonnt of work exjpended. And this explains why in the impact of imperfectly elastic bodies vis yiva is always lost, for a poition of the distortion remains. But ;io force is lost. One of the grandest generalizations of physical science is, that no energy in nature is lost. Iii the case of im- pact, compression develops heat, and this passes into the air or surrounding objects, and the amount of energy which is stored in the heat, electricity or other element or elements, which is developed by the compression, exactly equals that lost to the masses. We thus see that in the case of moving bodies, fo7'ce t7npels, momentu7n determines velocity after hnjpact^ and work or vis viva represents the resistance which tJie particles offer to heing displaced, 34. Statics is that case in which the force or forces which would produce motion are instantly arrested, resulting in pres- snre only. The express. .^n for the elementary work which a force can do is Fds^ but if the space vanishes, we have, Fds = 0. This, as we shall see hereafter, is a special case of "virtual velocities." The forces which act upon a body may be in equilibrium and yet motion exist, but in such cases the velocity is uniform. 35. The term power is often used in the same sense as force, but generally it refere to an acting agent. The term mechan- ical power is not onl}^ recognized in this science, but has a specific meaning, and for the purpose of avoiding ambiguity, it is better to use the term efo7't in reference to mechanical agents. Thus, instead of saying the power and weight, as is often done, say the effo7^t and resistaTwe. i8e-38.J INEETIA. 63 36. Inertia im/plies passiveness or want of power. It mcanB that matter has no power within itself to put itself in motion, or when in motion to change its rate of motion. Unless an external force be applied to it, it would, if at rest, remain for- ever in that condition ; or if in motion, continue forever in motion. Gravity, whi(ih is a force apparently inherent in mat- ter, can produce motion only by its action upon other matter. Inektia is not a force, but because of the property above explained, those impressed forces which produce motion are measured by the product of the mass into the acceleration as explained in preceding articles ; and many writei-s call this MEASURE the force of inertia. 37. Newton's Turee Laws of Motion Sir Isaac Newton expressed the fundamental principles of motion in the form of three laws or mechanical axioms ; as follows : — Ist. Every body continues in its state of rest or of uniform motion in a straight line unless acted upon by some external force. 2d. Change of motion is proportional to the force impressed, and is in the direction of the line in which the force acts. 3d. To every action there is opposed an ecpial reaction. [As simple as these laws appear to the student of the present daj^ the Bcience of Mechanics made no essential progress until they were recognized. See Whewell's Indtictive iSciencea^ 8d ed., vol. 1, p. 311. J 38. In all the problems thus far considered, it has beeD assumed that tlio ac- tion-line of the force or forces passed through the centre of the mass, prr)ducing a motion of translation only, l^ut if the action-line does not pass through the centre, it will produce both translation and rotation. p^^ ^^^ 64 DIRECT ECCENTRIC IMPACT. [38.) In eccentriG ir/ipact both translation and rotation is pro- duced. The centre of the hody will w.ove in a straight line^ hut every other jpoimt will describe arcs of circles in reference to the centre of the body^ which in space will be curves more or less elongated. The velocity of translation will be directly proportional to the intensity of the impulse imparted to the hody, but the angular velocity will depend upon the intensity of the impulse and the distance of the point of impact from, the centre of the body. In Figure 18, let Q = Mv be the impulse imparted to the body ; in which M is the mass of the body and v the velocity of the centre. Let this impulse be imparted at a. At h, a distance from the centre = cb = ac, let two eqnal and opposite impulses be imparted, each equal to iQ. The impulse Q, equals iQ -^ iQ. The four impulses evidently produce the same effect upon the body as the single impulse Q. If now one of the impulses, iQ, above the centre is combined with the equal and parallel one acting in the same direction below the centre, their effect will be equivalent to a single one, equal to Q applied at the centre c. This produces translation only. The other iQ above the centre combined with the equal and opposite iQ below the centre, produces rotation only; and it is evident that the greater the distance a, the point of impact, is from the centre, the greater will be the amount of rc^tation. An impact (or blow) at a to produce a velocity v at the centre of the body, must act through a greater space during contact, or the impacting body must move with a greater velocity, than if the impact be in a line passing through the centre c. (The entire energy stored in the body wili be -^Mv'^ 4- ilm<"^, in which Im is the moment of inertia of the rotating mass in reference to an axis through the centre, and w is the angular velocity in reference to the same axis ; and the other notation is the same as in the preceding Article. See Article 127. This expression for the energy, in case the bodies are perfectly elastic, will equal the energy lost by the impacting body. ) CHAPTER IL COMPOSITION AND RESOLUTION OF FORCES. OONOUKRENT FORCES. 39, If two or more forces act upon a material particle, they are said to be concurrent. They may all act towards the particle, or from it, or some towards and others from. 40, If several forces act along a material line, they are called conspirvng forces, and their effect will be the same as if all were applied at the same point. 41, The Resultant of two or more concurrent forces is that force which if substituted for the system will produce the eame effect upon a particle as the system. Therefore, if a force equal in magnitude to the resultant and acting along the same action-line, but in the opposite direction, be applied to the same particle, the system will be in equilib rrnm. If the resultant is negative, the equilibrating force will be positive, and vice versa. Hence, if several concurrent forces are in equilibrium, any one may be considered as equal and opposite to the resultant of all the others. 42, The resultant of several conspiring forces, eqicals the algebraic sum of the forces. That is, if i^ , i^ , i^ , etc., are the forces acting along the same action-line, some of which may be positive and the others negative, and li is the resultant ; then R = F^ + F^ + F^ + etc. = XF (45) 43, If two conxiurring forces be represented in m.agnitude and direction by the adjacent sides of a parallelogram,, the resultant will be represented in magnitude and direction by the diagonal of the 'parallelogram,. This is called the parol- Idogram of forces. If each force act upon a particle for an element of time it will generate a certain velocity. See equation (44). Let 5 PARALLELOGRAM OF FORCES. f4a^ the velocity which i^ would produce be represented b}^ AB ; and that of P by AD = BG. These represent the spaces >ver which the forces respectively would move the particle in a unit of time if each acted sepa- rately. If we conceive that the force F movea it from AtoB and that ' the motion is there ar- Fia. w. rested, and that P is then applied at i?, but acting parallel to AB^ then will the particle, at the end of two seconds, be at 0. If, next, we con- ceive that each force acts alternately during one-half of a second beginning again at A, the particle will be found at a in one-half of a second ; at J at the end of one second ; at c at the end of one and one-half seconds ; and finally at G at the end of two seconds. If the times be again subdivided the path will be Ad, de, ef^fb, hg, gh, hi, and iG, and it will arrive at G in the same time as before. As the divisions of the time increase, the number of sides of the polygon increase, each side becoming shorter; and the polj^gonal path approaches the straight line as a limit. There- fore at the limit, when the force P and i^act simultaneously, the particle will move along the diagonal, AG, of the parallelo-- gram. But when they act simultaneously, they will produce their etfcct in the same time as each when acting separately ; and hence, the particle w^ill arrive at G at the end of one second. Therefore, a single force B, which is represented by A G, will produce the same effect as P and F, and will be the resultant. If now a force equal and opposite to R act at the same point as the forces 7^ and P, the motion will be arrested and pressure only will be the result. See article 34. Hence, the parallelo- gram of velocities and of pressures becomes established."^ This is one of the most important propositions in Mechanics, and has been proved in a variety of ways. One work gives forty-five different proofs. A demonstration given by M. Poisson is one of the most noted of the analytical proofs. Many persons object to admitting the idea of motion in provinij th€ [i^.J TRIANGLE OF FORCES. 67 If ^ be the angle between the sides of the parallelogram which represent the forces P and i^ and li be the diagonal, or resultant, we have from trigonometry 12^ = 1^^ + 1^ + 2FF cos 6. (46) If 6 exceeds 90 degrees, it must be observed in the solution of problems that cos 6 will be negative. If ^ = 90 degrees, we have ^E Fia.2Q. >fR ->P Fio. 21 Also, if ^ = 90°, and a be the angle between B and P ; d fi between /i* and F\ then } (47) and fi between /i* and F\ then P — R cos a ; F = li {nm ^z=z R sin a. Squaring and adding, we have as before. /« -f i^ = R", The forces P and i^are called component forc£S. 44. Triangle of Forcbs. ff three concurrent forces are in eqaiUbrium^ they may he reprtHentcd in imigaitude and direc- tion by the sides of a triangle taken in their order ; and if the direction of action of one be revtrsedy it will be the resultanti of the other two. Tlins, in Fig. 19, if AB and BC lepresent two forces in magnitude and direction, ^6' will represent the resultant. parallelo^rram of prewurefl ; but we have seen that a prewnre when acting upon a free body will produce a certain amount of motion, and that this motion in a measure of the pressure, ami hence its use in the proof appears to be admissible. But the strongest proof of the correctness of the proposition i« the fact that in all the problems to which it has been a]>plied, the result! agree with those of experience and observation. es POLYGON OP FOE0B8. Since the sines of the angles of a triangle are proportional to the sides opposite, we have F _ P _ R m A sin P,R A sini^jff sin F^ POLYGON OP P0E0E8. 45. If several cor^ current forces are re- presented in magni- tude and direction hy the sides of a closed polygon taken in thei/r order, they will he i/n equilibrium. This may be proved by finding the resultant of two forces by means of the triangle of forces ; then the resultant of that resultant and another force, and so on. Fia 21a. PAKALLELOPIPED OF F0K0E8. 46. If three concurrent forces not in the same plane a/re represented in magnitude and direction hy the adjacent edges of a jparallelopipe- don, the resultant will he represented in magnitude and direction hy the diago- nal / and conversely if the diagonal of a parallelopipedon represents a force, it may he considered as the resultant of three forces represented hy the adjacent edges of the parallelopipedon. In Fig. 22, if AD represents the force Fi in magnitude and direction, and similarly I>B represents i^, and BG,F^\ J;hen according to the triangle of forces AB will represent the resultant of F^ and F^ ; and AG the resultant of AB and F^ and hence it represents the resultant of i^, i^, and F^ Fia.22. [48. J EXAMPLES. 60 If Fi, Fn, and F^ are at right angles with each other, we have Ii^ = F? -{- F? + F?; and if « is the angle B,Fi , |S of i?^ , and y of ByF^; then i^ = ^ cos o ; J i^ = ^ cos y. ) (49) Squaring these and adding, we have B^^F? + F^ + Fi\^ before. Examples. 1. When F^ i^ and ^ = 60°, find B ; (See Eq. (46) ). Ana. R = F\/Z. 2. n i^= i^i and ^ = 120°, find E. 3. If F=: i^ and ^ = 135°, find R. Ana, R = FV^^^Vf. 4. U F=2Fi = SR, find ^. 6. n |75"= f i^i = R, find the angle i^^Fi. Ana. 90° 6. If F=7,Fi = 9, and ^ = 25°, find ^ and angle f{r. 7. A cord is tied around a pin at a fixed point, and its two ends are drawn in different directions by forces i^^and P. Find 6 when the pressure upon the pin is ^ = ^ (P + F), Ana. cos 6 = ^p^ '- 8. When the concurring forces are in equilibrium, prove that F : F: Riisin F,R : sin F,R ; sin P^. 9. If two equal rafters suppoi-t a weight W at their upper ends, required the compression on each. Let the length of 70 CONCURRENT FORCES. [47.1 each rafter be a and the horizontal distance between their lower ends be b. Ans. V^aJ^-h^ W, 10. If a block whose weight is 200 pounds is so situated that it receives a pressure from the wind of 25 pounds in a due easterly direction, and a pressure from water of ]00 pounds in a due southerly direction ; required the resultant pressure and the angle which the resultant makes with the vertical. RESOLUTION OF OONOUREENT FORCES. 47* I^6t there be many concurrent forces acting upon a single particle, and the whole system be referred to rectangular co-ordinates. Let i^5i^, Fi, etc, be the forces acting upon a particle at A ; a?, y, z the co-ordinates of A ; «!, «2) etc., the angles which the Fio. 23. direction -lines of the respec- tive forces make with the axis of x ; /^i > ft > etc., the angles which they make with y ; yi J ^2 5 etc., the angles which they make with z ; and Z, y, and Z, the algebraic sura of the components of the forces when resolved parallel to the axes x, y, and s, respectively. Then, according to equations (45) and (49), we have for equilibrium ; X= i^cosai+7^cos«2 + ^cos «3 -h etc. =:: ^ Fcosa = 0;\ I^=i^iCOS|5i+7^cos^2 4-i<3C()sft + etc.=2'Fcos^= 0; V (50) Z=Fi cos yi+Fz c s j'2+^ cos y^ + etc. =-2:7^ cos y — 0)) If they are not in equilibrium, let It be the resultant, and by introducing a force equal and opposite to the resultant, the By stem will be in equilibrium. f48.] CONSTRAINED EQUILIBRIUM. 71 Let «, h and c be the angles which the resultant cnakes with the axes a;, y and z respectively ; then X = i? cos a ; r = i? cos J ; (51) Z = H COS c . Squaring and adding, we have X^+ T' + Z^ = I? (62) If jff = equations (61) reduce to (50). When the forces are in equilibrium any one of the JF'-fcrces may be considered as a resultant (reversed) of all the others. Equations (50) are therefore general for concurring forces. The values of the angles a, §, y, etc., may be determined b^ drawing a line from the origin parallel to and in the direction of the action oftheforce^ and measuring the angles from the axes to the line as in Analytical Geometry. The forces may always be considered as positive, and hence the signs of the terms in (50) will be the same as those of the trigonometrical func- tions. In Fig. 23 the line Oa is parallel to i^ , and the corre- sponding angles which it makes with the axes are indicated. If all the forces are in the plane x y then 71,^2? etc. = 90°, and (60) becomes X=-^i^cosa = 0; ) xKQx CONSTRAINED EQUTLIBRIUM. 48. A body is constrained when it is prevented from moving freely under the action of applied forces. If a particle is constrained to remain at rest on a surface under the action of any number of concurring forces, the lesultant of all the applied forces must be in the direction of the normal to the surface at that point. For, if the resultant were inclined to the normal, it could be resolved into two components, one of which would be pio. m. 72 CONSTRAINED EQUILIBRIUM, [48.> tangential, and would produce motion ; and the other normal, w^ich would be resisted by the surface. Let i\r= the normal reaction of the surface, which will be equal and opposite to the resultant of all the impressed forces ; ^aj = the angle {N'^x) ; By = the angle (iV,y) ; 6^ = the angle (iV",^) ; Z = (aj, y, z) = 0, be the functional equation of the surface; and JF[, F2, 1^3 9 etc., be the impressed forces. Then from (61) and (52), we have Z= iV^cos^,; (64) From Calculus we have COB^, = v/-(ir-(£y \dxl N/(f)'Hf)'M fdzy (55) and similarly for cos 6y and cos 6^ . These values in (54) readily give X _ r _ Z \dx) \dy J \d2 ) (66) After substituting the values of cos 0^, , cos 6^ , and cos 0^ in [48.1 CONSTRAINED EQUILIBRIUM. 78 (54), multiply the first equation by dx^ the second by «?y, the third by dz^ add the results, and reduce by the equation which is the total differential of the equation Z = ; and we have X.dx + Ydy -^ Zd2 = 0, (57) Equations (56) give two independent simultaneous equations which, combined with the equation of the surface, will deter- mine the point of equilibrium if there be one. Equation (57) is one of condition which will be satisfied if there be equilibrium. To deduce (55) let / («', y') = 0, and /' {af, e) = 0, be the equations of the normal to the surface at the point where the forces are applied. In Fig. 25 let Oa be drawn through the origin of oo-ordinates parallel to the required normal, then will dx\ dj/ and dz' be directly propor- tional to the co-ordinates of a ; w ,\ 000 aOsB = ooB 0x = g- dx' Via. 26. Vdx'*+ dy'^+dsf 1 But the normal is perpendicular to the tangent plane, and hence the pro- jections of the normal are perpendicular to the traces of the tangent plane. The Equation of Condition of Perpendicularity is of the form 1 -f oo' — (See Analytical Geometry) ; in which a' = zK^ , and a = ^ ; the latter of dcef da which is deduced from the equation of the surface ; ;-. 1 H- /,-/ =0; and similarhr dx dx ^ ^ ^ dz dz ^ do^ dm ' Y4 CONSTRAINED EQUILIBRIUM. r«. and -=-, = dL \ dz ) dz ~ 7dL\' the last terms of which contain the partial differential co-efficients deduced from the equation of the surface. These, substituted in the value of cos q^ above, and reduced, give equation (55). CONSTRAINED EQUILIBRITJM IN A PLANE. 49. If all the forces are in the plane of a curve, let the plane yx coincide with that plane ; then ^ = and (56) becomes •, idL\~ ldL\ \dx I \dy I (68) or, Xdx = — Ydy ; and, Xdx + Ydy = ; (59) in which the first of (58) may be used when the equation of the curve is given as an implicit function ; and the second of (58), or (59), when the equation is an explicit function. When the particle is not constrained it has three degrees of freedom (equations (50) ) ; when confined to a surface, two degrees (equations (56)) ; and when confined to a plane curve, only one degree (equation (58) ). Examples. 1. A body is suspended vertically by a cord which passes over a pulley and is attached to another weight which rests upon a plane; required the position of equilibrium. In Fig. 26, let the pulley be at the upper end of the plane and the cord and plane perfectly smooth. The Fi«. 86. weight P is equivalent to a force I4».l EXAMPLES. 75 which acts parallel to the plane, tending to move the weight W up it. Let W = the weight on the plane, which acts vertically downwards ; P = the weight suspended by the cord ; i = the inclination of the plane to the horizontal ; and Z=— y-faaj + 5 = 0, be the equation of the plane. Then X=Pcos^; T= -W + Fsmi; a = tan z sin t cos* (f) = -l'^°*'(S = »5 and these in (58) give P = TFsin*; which only establishes a relation between the constants, and thus determines the relation which must exist in order that there may be equilibrium ; and since the variable co-ordinates do not appear, there will be equilibrium at all points along the plane when P = TTsin i. The equation of the line, given explicitly, is y = ax + b; ,\dy = adx; which in the 2°** of (58), or in (59), gives, P = IF sin * as before. 2. Two weights P and TFare fastened to the ends of a cord, which passes over a pulley O ; the weight W rests upon a vertical plane curve, and P hangs freely ; required the position of equilibrium. The applied forces at IT are the weight TT, acting vertically downward ; the ten- sion P on the string ; and the normal reaction of the curve. [Consider the weight W and pulley as re- duced to points.] Take the origin of co-ordinates at the Fio. ft. 76 EXAMPLES. [49. J pulley, y vertical and positive upwards, and x positive to the right. The angle between + x and P is abc ; between + y and P, dfe. Let AW = x, OA = y, 0W= r, WOa =6; then sin ^ = ^ , cos 6 — —, r^ = x^ ■}- y^ ; X= F cos 270° -]- PGosabo=:0 + P cos (180° + 0) = - P cos 6, F= TFsin 270° + P sin (180° + d) = ^W- Psine; and (59) becomes ^ F cos e dx - {W + P sin 0) dy = 0; or .— ^ r^xdx + ydy j-.-, , . - Wdy = P ^— ^ = Pdr. (a) 3. Let the given curve be a parabola, in which the origin and pulley are at the focus, the axis vertical, and the vertex of the curve above the origin. The equation of the curve will be a^=z2p{-y + ip), in which 2j9 is the principal parameter. (See Analyt, Geom.) Differentiating gives xdx = —pdy, which substituted in (a) gives or which simply establishes a relation between constants; and therefore, if they are in equilibrium at any point they will be at every point, and there will be no equilibrium unless the weights are numerically equal. 4. Let the curve be a circle in which the origin and pulley are at a distance a above the centre of the circle. [49.] EXAMPLES. 77 Since y is negative downwards, we have for the equation of the circle, {a+yf + Q? = B?, Differentiating gives xdx = — (a + y) dy^ which substituted in {a) gives P 5. Let the curve be an hyperbola having the origin and pulley At the centre of the hyperbola, the axis of the curve being vertical. The equation of the curve will be ^^ — aV = a^b\ and if e be the eccentricity, we find hW y 6. Required the curve such that the weight TFmay be in equilibrium with the weight P at all points of the curve. This relation requires tliat the relation between y and x in equation {a) shall be true for all assumed values of P and W, We have Integrating gives -Wy -\- C=P V^T"^. Squaring or which is an equation of the second degree, and hence repre sents a conic. If P = TF", it is a parabola. If P > TF, it is an ellipse. If P < TF, it is a hyperbola. The origin is at the focus. 7. A particle is placed on the concave surface of a smooth sphere and acted upon by gravity, and also by a repulsive 78 EXAMPLES. [494 force, which varies inversely as the square of the distance from the lowest point of the sphere ; find the position of equilibrium of the particle. Take the lowest point of the sphere for the origin of coordi- nates, y positive upwards, and the eqr.ation of the surface will be Z = a? + y2 + s2- ^Ry = 0. Let r be the distance of the particle from the lowest point ; then /^ = ar»4.2/» + s2 = 2^y. (5) Let /A be the measure of the repulsive force at a unit's dis- tance ; then the forces will be ^ = -—^^ and mg = to = the weight of the particle. "^"^Ey'r' ^~UlyT "^^ ^ ^Ry r ' which in (56) give, after reduction, which in (5) gives, r^ = — R. To see if these values satisfy equation (57), substitute in it the values of X, Y, Z, and the final values of y and r, and we find, otxtx + ydy — Rdy + sd^ = ; which is the differential of equation (J), and hence is true. [This is the theory of the A^lectroscope.] 8. A particle on the surface of an ellipsoid is attracted by forces which vary directly as its distance from the principa planes of section ; determine the position of equilibrium. fOP, 51.] Let MOMENTS OP FORCES. Z = .^ (aj, y, 5^,) =-^ + -^ -f -^ -1= 0, 79 U" be the equation of the surface ; \dx)~a^' \dy)~0'' Kdz)"^' and let the x, y, and zcomponenta of the forces be rc8pct>> tively, X=-/tia?, Y=^it^^ Z-'^iL^; and (56) will give, which simply establishes a relation between the constants; and hence when this relation exists the particle may be at rest at any point on the surface. The result may be put in the form, /^i _ /^ __ /*8 _ Ml +/^ 4- /is ly' cr^ "" a-» 4- h-'' 4- c^ MOMENTS OF F0R(^E8. 50. Def. The Tnoment of a force in reference to a point is the product arising from nudtiplying the force hy tfie perpendicular distance of the action-llTie of tlie force from the point. Thus, in Fig. 2S, if O is the point from wliich the perpendicular is drawn, F the force, and Oa the peq^endicular, then the moment of F\% EOa = Ff', ID which y is the perpendicular Oa, Fio. S8. 51, Nature of a momknt. The moment of a force measures the turning or twisting effect of a force. Thus, in Fig. 28, if the particle upon which the force F acts is at .4, §0 MEASUBE OF A MOMENT. [6»-54.J and if we conceive that the point is rigidly connected to J., the force will tend to move the particle about O^ and it is evident that this effect varies directly as F. If the action- line of F passed through it would have no tendency to move the particle about that point, and the greater its dis- tance from that point the greater will be its effect, and it will vary directly as that distance ; hente, the measure of the effect of a moment varies as thejprodMct of the force and jperpendi- cular / or as oFf; where o is a constant. But as c may be chosen arbitrarily^ we make it equal to unity, and have simply Ff as given above. 52. Def. The point O from which the perpendiculars are drawn is chosen arbitrarily, and is called the origin of mo- ments. When the system is referred to rectangular coordinates, the origin of moments may, or may not, coincide with the origin or coordinates. The solution of many problems is simplified by taking the origin of moments at a particular point. 53. The lever arm, or, simply, the arm, of a force is the perpendicular from the origin of moments to the action-line of the force. Thus, in Fig. 29, Oa is the arm of the force F^ ; Og that of the force F^ , etc. Generally, the arm is the per- pendicular distance of the action-line from the axis about which the system is supposed to turn. 54. The sign of a moment is considered positive if it tends to turn the system in a direction opposite to that of the hands of a watch; and negative, if in the opposite direc- tion. This is arbitrary, and the opposite directions may be chosen with equal propriety ; but this agrees with the direction in which the angle is computed in plane trigonometry. Gen- erally we shall consider those moments as ^positive which tend to turn the system in the direction indicated by the nat- ural order of the letters, that is, positive from + a? to ■\- y ; from + ^ to -h 3 ; then from + s to -f- a? ; and negative in the reverse direction. The value of a moment may be represented by a straight line drawn from the origin and along the line about which T55-M.J MOMENT OF A FORCE. 81 rotation tends to take place, in one direction for a positivB value^ and in the opposite direction for a negative one. 66. The composition and resolution of moments may be effected in substantially the same manner as for forces. Thej may be added, or subtracted, or compounded, so that a resultant moment shall produce the same effect as any num. ber of single moments. The general proof of this proposi- tion is given in the next Chapter. 66. A MOMENT AXIS is a line passing through the origin of moments and pei-pendicular to the plane of the force and arm. 67. The moment of a fobgb kefebred to a moment axis is the product of the force into the perpendicular distance of the force from the axis. If, in Fig. 29, a line is drawn through O perpendicular to the plane of the force and arm, it will be a moment axis, and the turning effect of Fl upon that axis will be the same wher- ever applied, providing that its arm Oa remains constant. If the force is not perpendicular to the arbitrarily chosen axis, it may be resolved into two forces, one of which will be perpendicular (but need not intei-sect it) and the other parallel to the axis. The moment of the former component will be the same as that given above, but the latter will have no mo ment in reference to that axis although it may have a moment in reference to another axis perpendicular to the former. 58. The MOJkfENT OF . A FORCE IN REFERENCE TO A PLANE TO WHICH IT IS PARALLEL is the product of the force into the distance of its action-line from the ylane. 69. If dny numher of concurring forces are in equilibrium the algebraic sum of their moments will he zero. Let Fi,Fi,Fs, etc.. Fig. 29, be the forces acting upon a particle at A ; and O the assumed origin of moments. Join and A^ and let fall the per- pendiculars Oa^ Oh, Oc, etc., upon the action-lines of the respective forces, and let Oa=f\ Oh=f^\ Oc=f\ etc. fio. 29. 92 RESULTANT MOMENT. r6u.j Eesohe the forces perpendicularly to the line OA ; and since they are in equilibrium, the algebraic sum of these c^)m- ponents will be zero ; hence, Fi sin OAIl -\-Il sin OAF,, + F^ sin OAF<, + etc = ; Oa OA Ob Oc or, F^^:^ + i? ^ + i^s^ + etc. = 0, Multiply by OA, and we have FiOa + F^Ob + F^Oc + etc. = 0; or, i^/i + i^/a + i^/3 + etc. = 2Ff^ 0. (60) It is evident that any one of these moments may be taken as the resultant of all the others. MOMENTS OF CONCUERING FORCES WHEN THE SYSTEM IS BEFEHBED TO RECTANGULAR AXES. 60. Let A, Fig. 30, be the point of application of the forces i^, j?2, i^, etc., and O the ori- gin of coordinates, and also the origin of moments. Let a?, y, and z be the coordinates of the point A. Resolving the forces parallel to the coordinate axes, have, from equation (50), B c y f« / \ y,r, ^ \4 L / / we Fig. 80. X= ^i^cos a; Y=2.Fito%^\ Z = 2F(io%ry, The X-forces prolonged will meet the plane of yzmB", and will tend to turn the system about the axis of y, in reference to which it has the arm^6^ = ;s; and also about s, in refer- ence to which it has the arm BD = y. Hence, employing the notation already established, we have for the moment of the sum of the components parallel to a?, — Xy^ and -f Xz. f(K).] MOMENTS OF CONCURRING FORCES. 88 Similarly for the y-GOinponents we find the moments, -h Yx, and — Yz ; and for the z-comjponents, — Zr, and + Zy. The moment Xy tends to turn the system one way about the axis of s, and Yx tends to turn it about the same axis, but in the opposite direction ; and hence, the combined effect of the two will be their algebraic sum ; or Yx—Xy. But since there is equilibrium the sum will be zero. Com- bining the others in the same manner, we have, for th4 moments of cancurrmg forces, in equilibrium: In reference to the axis of a? . * . . . Zy — Yz = 0\ . Xz — Zoj = (61> " " " " " " 3 . . . , Jo? - Xy== 0. The third equation may be found by eliminating z from the other two ; hence, when X, Y, and Z are known, they are the equations of a straight line ; and are the equations of the resultant. If the origin of moments be at some other point, whose coordinates are x' , y\ and z' ; and the coordinates of the point A in reference to the origin of moments be zji, yi,and Zi\ then will the lever arms be x^=^x-x' \ yx = y — y'\ When the system is referred to .rectangular coordinates the arm of the force, referred to the s-axis, is X cos /3 — y cos a, in which y and x are the coordinates oi any point of the action-line of the force ; and a is the angle which the action-line makes with the axis of a?, and y8 the angle which it makes with y- Fio. 81. 84 EXAMPLES. [60.1 In Fig. 31, let ^i*^ be the action-line of the force i^ O the origin of coordinates, A anj^ point in the line AF^ oi which the coordinates x = Ob, and y = Ah. Draw Od and he perpendicular to AF, and hd from h parallel to BF, The origin of moments being at 0, Oa will be the arm of the force. We have Agh = a = cbA^ cAb = ^z=zhOd, cb=:y cos a = ad, Od = X cos l3; .•. Oa = Od — ad— xcosfi ~ 2/ cos oc. {61a) If there are three coordinate axes, this will be the arm in reference to the axis of b ; and if there be many forces, the Bum of their moments in reference to that axis, will be SF(x cos 13 — y cos a). Examples. 1. A weight W is attached to a string, which is secured at A, Fig. 32, and is pushed from a vertical by a strut OB ; required the pressure i^on BO when the angle GAB is B, The forces which concur at B are the weight IT, the pressure F, and the tension of the string AB, Take the origin of moments at A^ and we have - W.BC^- F,AO+ tension x = ; . Fxo. 8S. 2. A brace, AB, rests against a vertical wall and upon a horizontal plane, and supports a weight W at its upper end ; required the compression upon the brace and the thrust at A when the angle OAB is 6 ; the end B being held by a string B C. Fxo.sa [90,] EXAMPLES. 85 The concurring forces at A, are W, acting vertically down- ward, the reaction of the wall JV acting horizontally, and the reaction of the brace K Take the origin of moments at ^, we have -mDB+ W.CB+F.O=0\ .•.iV^ = >f tan^. Taking the origin of moments at D, we have TT. AD-F. i?^ sin ^ + iT. = ; .',F= TTsec^. 3. A rod whose length is BC= I is secured at a point By in a horizontal plane, and the end O is held up by a cord AC 80 that the angle BAOis 6, and the distance AB = a ; required the tension on AO and compression on BC, dne to a weight TT applied at O, Ana, CoTwpresdon — — TTcot Q, 4. A cord whose length ABC = I is secured at two points in a horizontal line, and a weight W is suspended from it at B ; required the tension on each part of the cord. /^ CHAPTER m. PARALLEL F0B0E8. 61, Bodies are extended masses, and forces may be applied at any or all of their points, and act in all conceivable direc- tions, as in Fig. 34:. 62. Suppose that the action- lines OF ALL the forces ARE PARAL- LEL TO EACH OTHER. Tliis is a Spe- cial case of concurrent forces, in wliicli the point of meeting of the action-lines is at an infinite distance. la Fig. 35, let the points a, J, G, etc., which are on the action- lines of the forces and within the body, be the points of applica- tion of- the forces, and the point where they would meet if pro- longed. If the point O recedes from the body, while the points of application a, b, c, etc. remain fixed, the action-lines of the forces will approach parallelism with each other, and at the limit will be parallel. Fzo.84 Fia. 85. 63, Resultant of parallel forces. The forces being par- allel, the angles which they make with the respective axes, including those of the resultant, will be equal to each other Hence, I64.J MOMENTS OF PARALLEL FORCES. 97 a = cii =^ a^ = Oq, etc. "^ a ; 5 = /3i = i8a^/88,etc. = iS; c = 7i = 72 =78, etc. =7; and those, in equations (50) and (51), give X= Bco&a= (Fi + Fi+ Fs + etc.) cos a. Y= B cos ^ = (Fi ^ Fi -h F^ + etc.) cos /9. (62) Z=7?cos7 = (i^ + i^ + i^+ etc.) cos 7. From either of these, we have Ii = Fi-\-Fi-¥F9+ etc. = SF (63) Ilence, the resultant of j>aralld forces eqtuds the algebraid sum of the forces. From (62), we have which is the same as (52). MOMENTS OF PARALLEL FORCES. 64, Let Fi, Fi, F^, etc., be the forces, and aji, yi, Si ; aj„ yi, 029 etc., be the coordinates of the points of application of the forces respectively (which, as before stated, may be at any point on their action-lines). Then the moments of Fi will be, according to Article (60), in reference to the axis of «, j?i cos 7 . y^ — ^ cos )9 . ^i ; " " " '' " " y,i^co8a.2i -7^cos7.jBi; u a « « « " 5, i^co8)9.aJi-i^iCOsa.yi; and similarly for all the other forces. Ilence, the sum of the moments in reference to the respective axes for equUibri im is, ho, (^1 + Fzi/i + Fsi/s + etc.) cos 7 —{FiSi + ifej + ifej + etc.) cos 13 (FiSi + F222 + FsZs + etc.) cos a -^{FiXi + F2PC2 + FsX^ + etc.) cos y I- (FiSi + 7^ + ife + etc.) cos a ) ^ = 0; {FiXi + F^ + F^ + etc.) cos iS; -(-%! + F^yi + F^8 + etc.) cos a 88 THREE PARALLEL FORCES. [65.| These equations will be true for all values of a, /S, and 7, if the coefficients of cos a, cos ^, cos 7, are respectively equal ta» zero ; for which case we have Fx^h. + ^aj2 + i^a?3 + etc. = "ZFx = ; ■^yi + ^2 + -^8 + etc. = SFy = ; i^2i + i^32 -1- i^Ss + etc. = SF2 = ; (64) from which the coordinates of the point of application of anjr one of the forces, as i^, for instance, may be found so as ta satisfy these equations, when all the other quantities are given. Let the given forces have a resultant. Let a force, as i^, equation (64), equal and opposite to the resultant, be introduced into the system, then will there be equilibrium. Let SFx, SM/y XFz^ include the sum of the respective products for all the forces except that of the resultant; B be the resultant, and x^ y^ ^, the coordinates of the point of application of the result- ant ; so chosen as to satisfy equations (64), then we have Bx-XFx=0'^ Ry-XFy = 0'^ Rj-XFz^O. (65) Substitute the value of i? = SF, in these equations, and we- find _ XFx _ XFy _ ^Fb ,__, by which the point of application of the resultant becomes known, and, being independent of a, yS, and 7, is a point through which the resultant constantly jpasses, as the forces are turned about their jpoints of apjplication^ the forces co7istantly retaining thei/r jparallelism. This point is called the centre of parallel forces, 65. Is* THE SYSTEM CONSISTS OF THREE FOJRCES ONLY, and are- in the plane icy, we have R=.Fx + F^:, Fxx, + F,x^ ^ .-^_ F^,-^F^, [ {67} ces.J STATICAL COUPLES. 89 kF^ FlO. 86. Ist. Consider Fx and Flas positive, ^ The resultant will equal the arith- metical sum of the forces. Take jc* the origin at a Fig. 36, where the resultant cuts the axis of a;; then a = 0, and the second of (67) gives T^a^ = - i^ ; and hence, if i^ > i^, avj will exceed oJi ; that is, the resultant is nearer the greater force. 2d. Consider Fz a^ ne^atim. ^F. In this case the resultant equals the difference of the forces. Take the origin at a, Fig. 37, and we have FxXr = F^', and hence both forces are either at the p, right or left of the resultant. 3d. Let Fi = Fi=^ Ff and one of the forces be negative, then E = F-F=0; x^ ^p^^^ =y^ ; andy = oo ; (68) that is, the resultant is zero, while the forces may have a finite moment equal to F{xi± x^. Such systems are called 1 Pio. 87. COUPLES. 66. -^ GOV{ple consists of two equal parallel forces acting in opposite directions at a finite distance from eoxih other, A statical couple cannot be equilibrated by a single force. It does not produce translation, but simply rotation. A couple can he equilibrated only hy an equivalent couple. Equi/oalent couples are such as have equal moments. TUe resultant of several couples is a single couple which will produce the same effect as the component couples. 90 AXIS OF A COUPLK [67- «8.] 67» The arm, of a couple is the ^erpendicula/r iistartce hetween the action-lines of th^ forces. Thus, in Fig. 38, let O be the origin of coordinates, and the axis of x per- pendicular to the action-line of F\ — X then will the moment of one force be Fxi , and of the other Fx^ , and hence the resultant moment will he Xz F Xi F Fio.88. F{x,-X2) = Fa^; (69) hence, ah is the arm. If the origin of coordinates were between the forces the moments would he F (xi + x^) = F.ah as before. If the origin be at a we have FS) -f F.ah = F.ah as before. 68. The axis of a statical couple is any line joerpen- dicular to the plane of the couple. The length of the axis may be made proportional to the moment of the couple, and placed on one side of the plane when the moment is positive, and on the opposite side when it is negative ; and thus com- pletely represent the couple in magnitude and direction. If couples are in parallel planes, their axes may be so taken that they will conspire, and hence the resultant couple equals the algebraic sum of all the couples. If the planes of the couples intersect, their axes may intersect. Let O = F.db = the moment of one couple ; 6>i= Fi.aJ)i = the moment of another couple ; 6 = the angle between their axes ; and Os = the resultant of the two couples ; then and this resultant may be combined with another and bo on until the final resultant is obtained. Examples. 1. Three forces represented in magnitude, direction and position^ by the sides of a triangle, taken in their order, produce a couple. EXAMPLES. 91 2. If three forces are represented in magnitude and position by the sides of a triangle, but whose directions do not follow the order of the sides ; show that they will have a single resultant, 3. On a straight rod are suspended several weights ; Fx = 5 lbs., i^ = 15 lbs., /; = 7 lbs., /; = 6 lbs., 7i? = 9 lbs., at dis tanccs AB = 3 ft., BD = 6 ft., DE=5 ft, and £:F= 4 ft. ; required the distance AO at which a fulcrum must be placed so that the weights will balance on it ; also required the pressure npon it. 4. The whole length of the beam of a false balance is 2 feet 6 iiH hes. A body placed in one scale balances 6 lbs. in the other, but when placed in the other scale it balances 8 lbs. ; required the true weight of the body, and the lengths of the arms of the balance. 5. A triangle in the horizontal plane x, y has weights at the •everal angles which are proportional respectively to the opposite sides of the triangle ; required the coordinates of the centre of the forces. Let iCi , yi be the coordinates of A, X2,y^ofB; x^.y^oi C\ X, y of the point of application of the ro^oltant ; then we have (a + & + c)S=aiCi + JiC2-f«zj8; and (a + 5 + c) y = ayi + ly^ + cy^, • 6. If weights in the proportion of 1, 2, 3, 4, 5, 6, 7 and 8 arc Buspcndod from the respective angles of a parallel opiped ; re- quired the point of application of the resultant. 7. Several couples in a plane, whose forces are parallel, are applied to a rigid right line, as in Fig. 40 ; required the re- Bultant couple. i d V. V, r, rza. 4Q» CENTRE OF GRAVITY [69. J 8. Several couples in a plane, whose respective arms are not parallel, as in Fig. 41, act upon a rigid right line ; required the resultant couple. Fio. 41. CENTRE OF GRAVITY OF BODIES. 69. The action-lines of the force of gravity are normal to the surface of the earth, but, for those bodies which we shall here consider, their convergence will be so small, that we may consider them as parallel. We may also consider the force as the same at all points of the body. The centre of gravity of a hody is the point of application of the resultant of the force of gravity as it acts upon every particle of the body. It is the centre of parallel forces. If this point be supported the body will be supported, and if the body be turned about this point it will remain constantly in the centre of the parallel forces. Let M = the mass of a body ; m = the mass of an infinitesimal element ; . V = the volume of the body ; D = the density at the point whose coordinates are a?, y^ and z ; jK = TT = the resultant of gravity, which is the weight ;, and a, y, and i be the coordinates of the centre of gravity. We have, according to equations (63) and (20), E=^W=Xgm = M X g\ and (65) becomes X Sg^n = 'Xgmx ; or Mx = ^mx ; ) y Sgm = ^gmy ; or My = Smy ; V (70) 2 Sgm = Xgmz ; or Jfi = 'Srm, ) If the density is a continuous function of the coordinates of lihe body we may integrate the preceding expressions The- C».l OF BODIES. 98 complete solution will sometimes require two or three integra- tions, depending upon the character of the problem ; but, using only one integral sign, (22) and (70) become ^/DdV=fnxdV\ y/DdV=fDydV', z/BdV^/DzdV, (71) If the origin of coordinates be at the centre of gravitj, then and hence, Xmx=fl)xdV-0'^ and similarly for the other values. If 2> be constant, this becomes f,','xdY=0', (71a) {71J) the limits of integration including the whole body. If the mass is homogeneous, the density is uniform, and J) being cancelled in the preceding equations, we have - fxdV y — -^-y > z 1^. (72) Many solutions may be simplified by observing the following principles : 1. If the hody haa cm axis of symmetry the centre ofgramby xmU he on that axis. 94 CENTRE OF GRAVITY [7a ) 2. If the hodyhas a plane of symmetry the centre of gravity vyill he in thatjplane, 3. If the hody has two or more axes of symmetry the centrr of gravity will he at their intersection. Hence, the centre of gravity of a physical straight lino of uniform density will be at the middle of its length ; that of the pircumference of a circle at the centre of the circle ; that of the circumference of an ellipse at the centre of the ellipse ; of the area of a circle, of the area of an ellipse, of a regular polygon, at the geometrical centre of the figures. Similarly the centre of gravity of a triangle will be in the line joining the vertex with the centre of gravity of the base ; of a pyramid or cone in the line joining the apex with the centre of gravity of the base. There is a certain inconsistency in speaking of the centre of gravity of geometrical lines, surfaces, and volumes ; and when tliey are used, it should be understood that a line is 2ijohysical or material line whose section may be infinitesimal ; a surface is a material section, or thin plate, or thin shell ; and a volume is a mass, however attenuated it may be. When a body has an axis of symmetry, the axis of x may be made to coincide with it, and only the first of the preceding equations will be necessary. If it has a plane of symmetry, the plane x y may be made to coincide with it, and only the first and second will be necessary. 70, Centre of gravity of material Ivnes^ Let & = the transverse section of the line, and ds = slu element of the length, then dV=^kd8; and (71) becomes xfDMs = fDhcds ; yfDMs = fDhyds ; \ (73) zfDUs = fDksds, |70.1 OF LINES. If tLe transverse section and the density are uniform, we have » = -^: ^=•^5 -z=J^ (74) The centre of gravity will sometimes be outside of the line or body, and hence, if it is to be supported at that point, we must conceive it to be rigidly connected with the body by lines which are without weight. Examples. 1. Find the centre of gravity of a straight fine wire of uni- form section in which the density varies directly as the distance from one end. Let the axis of x coincide with the line, and the origin be taken at the end where the density is zero. Let 5 be the density at the point where a? = 1 ; then for any other point it will be D — hx\ and substituting in the first of (73), also making <fo = ^, we have xdx — I Saj^dx; ^0 x=ia. This corresponds with the distance of the centre of gravity of a triangle from the vertex. Vie. 42. 7io. 4S. t 8 Find the centre of gravity of a cone or pyramid, whether 96 EXAMPLES. [70.1 right 01 oblique, and whether the base be regular or irre- gular. Draw a line from the apex to the centre of gravity of the base, and conceive that all sections parallel to the base are re- duced to this central line. The problem is then reduced to finding the centre of gravity of a physical line in which the density increases as the square of the distance from one end. Ans, X = ia. 3. To find the centre of gravity of a circular arc. Let the axis of x pass through the centre of the arc, -5, and the centre of the circle 0; then F= 0. Take the origin at -5 : and let then, x = BI}, 25 = the arc ABC, and r = OG = the radius of the circle ; y^ = ^rx — a^. Differentiate, and we have ydy = rdx — xdx ; dy r—x dx . . . ds = — which = — ; y hence, the first of (74:) gives Jo ^ xdx V 2rx — a^ X = [ V 2rx — a? + s Jo « / ry\ ry which is the distance Bo. Then gO = r — ir -)= — » V 8/8 hence, the distance of the centre of gravity of an arc fr^m the {71. J CENTRE OF GRAVITY OF SURFACES. 97 centre of the circle is a fourth proportional to the aic, the radius, and the chord of the arc. 71. Centre of gravity of plane surfaces. Let the coordinate plane xy coincide with the surface ; thcD dV = docdy-y .'. V = ffdosdy = fydx or fosdy ; and (Tl^l be- comes xffjDdxdy=ffDxdxdy; ) yffI)dxdy=/fDydxdy, \ ^^^^ The integrals are definite, including the whole area. The order of integration is immaterial, but after the first integra- tion the limits must be determined from the conditions of the problem. If D is constant and the integral is made in respect to y, we have -^fy^^ fydx' \ffd^ (76) ^ fydx and if x be an axis of symmetry, the first of these equations will be sufficient. If the surface is referred to polar coordinates, then dV ■=. pdpdd, and x = p cos d^y z=z p sin ^, and (71) becomes __ ffPp^o^edpdd "^ - ffvpdpde «_ ffPphmedpde y-' ffDpdpdd (T7) Examples. 1. Find the centre of gravity of a semi-parabola whoso equation is 2/^ = 2j9a?. Equations (76) become r» V'ijp^ dx '= fi = »•! /:< ^ '^xdx 9? CENTRE OF GRAVITY. {72.|. 2. Fiiid the centre of gravity of a quadrant of a circle m which the density increases directly as the distance from the centre. Let 8 = the density at a unit's distance from the centre ;. then I) = 8p at SL distance p ; and (77) becomes hi I p^cosedpdd Vo Jo '''''' 72, Centre of gravity of curved surfaces. We have for an element of the area dY^=-dxdy x sec, of the angle hetween the tangent plane and' the coordinate plane xy ; or, d V=sec, 6xdx dy; at, in terms of partial differential coefficients ,/ (dL\^ 7dL\' ldL\» ''=JJ fA ^^' m and, for a homogeneous surface, (72) becomes as = — y— , i — — p=~ ' « F~' J w, the surface may be referred to polar coordinateik (78). 172.] OP DOUBLE CUBVED 8X7BFACBS. »9 If tlie surface is one of revolntion, let x coincide with the axis of revolution, then x/nyds ^zfnyxda. (78a) Example. Find tue centre of gravity of one-eighth of the surface of • sphere contained within three principal planes. Let the equation of the sphere be Z = ar» + 2^4-38-^ = 0; then dL __ dL ^ dL dx~ ^ di and the first of (78) becomes dx ^^ dy ^^ dz 2 np X dx dy 2 /• -— . -n* = 'vJ^-jf "' = ^n? ^^y V ^^lyVW^-^IPsin-^g^ Similarly, y = ili = i. This problem may be easily solved by the aid of elementary geomotxy, Conceive that the snrface is divided into an indefinite number of small zones by equidistant planes which are perpendicular to the axis of 2, in which 100 CENTRE OF GRAVITY OF VOLUMES. [78.1 the area of the zones will be equal to each other. Conceive that these zonea are reduced to the axis of x ; they will then be uniformly distributed along that axis, and hence the centre of gravity will be iB from the centre ; and aa the surface is symmetrical in respect to each of the three axes, we get th« Bame result in respect to each. 73, Centre of gravity ofvolwnea, or heavy bodies. We have dV = dxdyds; /. xjyfD dx dyd3=JXf Dx dx dy dz ; (79) and similarly for y and i. If aj is an axis of symmetry, (79) is suflScient. If the surface is referred to polar coordinates, let ^ = AOx, e = dOA, then, gd = dp, gf= pd0, gh = p cos 6d<l), and, dV= p''dpco8dded<t>, ^ Fia. 45. also, X = pco8 6 co8<l); y = p sin 6 ; and 3 = poos dsvn^. Hence, for a homogeneous body, we have Vx = jy/p^ Gos^ d cos <l> d p d e d (j)]^ Vy ^fff p^cosQ sm>edpded<^\ \ (80) YT^f/fp^cos'esin(tidpded<\>. j If the volume be one of revolution about the axis of a?, i have [73.] EXAMPLES. 101 Yx=nffxdxA ^^^' Example. Find the centre of gravity of one-eighth of the volume of a homogeneous ellipsoid, contained within the three jprincvpal planes. Let the equation of the ellipsoid be -, + ^ + ^-1 = 0; then, equation (79) gives, /»6' pY pX ro rY nX X I I I dxdifdz— I I I xdxdydz. Jo Jo Jo Jo Jo Jo Performing the integration, we have \ iraho x = ^ ird^bo ; /. i = f a. Similarly, y = |J, and z = |c. Performing the above integration in the order of the letters rv, y and f , and nsing the limits in the reverse order as indicated, we have for th« X'limita. and the corresponding limits for y will be y = bK/l — ^ = Y; &adi/ = 0. For Hiefirtt member of the equation, we have Y /.X Consider kJ \ —*-^ — B, a oooftant in ref erenoe to the y-inUgratttm and we have Gonsider as constant in performing the y-integraUony and we have T Ml v z*?y \ _ I 102 EXAMPLES. f74l For the second member, we haye Q performing the y-integration^ i •i\-*-'Ml ^vdbc ^ as given above. 74. When the centre of gravity of a body is known and the centre of gravity of a part is also known, the centre of the remaining part may be found as follows : — Let W = the weight of the whole body ; X = the distance from the origin to the centre of gravity of the body ; Wi = the weight of one part ; Xi = the distance of the centre of w^ from the ftame origin ; W2 = the weight of the other part ; and aS| = the distance of the centre of w^ from the same origin ; then w^Xi + w^ = {wi + w^x= Wx; ftnd hence, Wx — WiXi ,^- ^75. THEOREMS OF PAPPUS. 103 If the body is homogeneous, the volumes may be s ibstitttted for the weight. Example. Let ABO be a cone in which the line ME joins the vertex and the centre of gravity of the base; and the cone ADG^ having its apex Z>, on the line BE, and the same base ^ 6^ be taken from the for- mer cone, required the centre of gravity ^f the remaining ])artj AD OB. Let F = the volume of ACB, a = BE, (h = BE, X = Ec ■=^ ia — the distance of the centre of ABO from E, jCi = Ed = iai= the distance of the centre of ABO from E; then, Fia. 46. «1 V = the volume of A OB, and (82) becomes (h ajj = F. i « - F. ^. i(h a F- F^ a = i (a + ai). Centeobarig Method, or 75, Theorems of Pappus or of Guldintjb. Multiply both members of the second of (74) by 27r, and it Tiiay be reduced to 27rys = lirfyds, (83) the second member of which is the area generated by the revo- lution of a line whose length is % about the axis of a?, and the 104 EXAMPLES. I75.t first member is the circumference described by the centre oi gravity of the line, multiplied by the length of the line ; hence, the area generated hy the revolution of a line about a fixed axis equals the length of the line multiplied hy the drcurnfer' ence described by the centre of gravity of the line. This is one of the theorems, and the following is the other. From the second of (76), we find 27ry A —f'K'fdx. The right-hand member, integrated between the proper limits, is the volume generated by the revolution of a plane area about the axis of x. The plane area must lie wholly on one side of the axis. In the first member of the equation, A is the area of the plane curve, and 27ry is the circumference described by its centre of gravity. Hence, the volume gener- ated by the revolution of a plane curve which lies wholly on one side of the axis^ equals the area of the curve multiplied hy the circumference described by its centre of gravity. Examples. 1. Find the surface of a ring generated by the revolution of a circle, whose radius is r, about an axis whose distance from the centre is c. Ans. 4fn^rc. 2. The surface of a sphere is 47r7^, and the length of a semicircumf erence is tt r ; required the ordinate to the centre of gravity of the arc of a semicircle. 3. Required the volume generated by an ellipse, whose semi- axes are a and J, about an axis of revolution whose distance from the centre is c ; c being greater than a or h. (Observe that the volume will be the same for all positions of the axes a and h in reference to the axis of revolution.) 4. The volume of a sphere is ^ir /^, and the area of a semi- circle is ^ /^ ; show that the ordinate to the centre of gravift _ 4r of the semicircle is y = «— . [7«.J ' EXAMPLES. 105 76. Additional Examples. 1. Find the centre of gfravity of the quadrant of the circumference of a drcle contained between the axes x and y^ the origin being at the centre. . - 2r - Aim, X = — = y. IT 2. Find the distance of the centre of gravity of the arc of a cycloid from the Tertex, r being the radius of the generating circle. Ans. y = |r. 3. Find the centre of gravity of one-half of the loop of a leminiscate, of which the equation is r' = a' cos 20, I being the length of the half loop. rt2 _ a* ^2 - 1 Ans. X = — -; y = -J 7=^. 4 Find the centre of gravity of the helix whose equations are X = a cos <p^ y = a Bin (p^ z = na (pf the helix starting on the axis of x. — V — o> — X - Ans. X = na- '^ y = na ; and e = i*. z z 6. Find the centre of gravity of the perimeter of a triangle in space. 6. If Xo and yo are initial points of a curve, find the curve such that mas = X — Xoj and ny — y — y^ 7. A curve of given length joins two fixed points ; required its form so thai itB centre of gravity shall be the lowest possible. (This may be solved by the Calculus of Variations). An; A Vatenary, 8. Find the centre of gravity of a trapezoid- Let ADEB be the trapezoid, in which DE and ,f./^ AB are the parallel sides. Produce AD and BE / j \ until they meet in O, and join G with F, the mid- /' / \ ile point of the base ; then the centre of gravity will be at some point g on this line. The centre of the j triangle ACB will be on CF, and at a distance of / kOF from F; and simUarly that of JDCE wUl be ^ / on the same line, and at a distance of i CG from -^ a ; then, by (82) we may find Fg. ^"- ^'^- f/ I Ans, Fg = kFQ AB + 2DE AB -r DE ' (If DE is zero, we have Fg = \FC for the centre of gfravity of the triangle ABC.) 9. Find the oentre of giarit^ of the quadrant of an ellipse, whose equatioi» Am. «=i-; jr = f:;;- 106 EXAMPLES. [7«.l 10. Find the centre of gravity of the circolai sector AJBC. Let the angle ACB =26; then _ , r sin d 11. Find the centre of gravity of a part of a circular annulus ABED, Let AC = r, 1)0 = n, and ACB =2d ; then sin e r^ + TTi + rj» Ans. Cgi = J 12. Find the centre of gravity of the circular spandril FOB. 18. Find the centre of gravity of a circular segment. Ans. Dist from C = r+r, {cTuyrd)^ 12 area of segment 14. Find the distance of the centre of gravity of a complete cycloid from its vertex, r being the radius of the generating circle. Ans. y = Ir. 15. Find the centre of gravity of the parabolic spandril OCB, Fig. 49, in which DC = y, and CB = x. _ An^. X = ^x; 16. Find the centre of gravity of a loop of the leminiscate, whose equation is r' = a' cos 29. Ans. x = —z a. Fig. 49. 17. Find the centre of gravity of a hemispherical surface. Ans. x = ir. 18. Find the centre of gravity of the surface generated by the revolution of a semi-cycloid about its base, a being the radius of the generating circle and J the distance from the vertex of the surface. Ans. aj=r|a. 19. The centre of gravity of the volume of a paraboloid of revolution ia X = ix. 20. The centre of gravity of one half of an ellipsoid of revo'ution,of which the equation is a'y' + b^x^ = a^b^y is « = fa. 81. The centre of gravity of a rectangular wedge ia » = ta. [78] EXAMPLES. 107 22. The centre of gokritj of a semicircular cylindrical wedc^, whose radina tor, ii Fia. 60. Fia. SI. 28. The vertex of a right circular cone is in the surface of a sphere, the mxis of the cone passing through the centre of the sphere, the base of the cone being a portion of the surface of the sphere. If 29 be the vertical angle of the •one, required the distance of the centre of gravity from the apex. . 1 — cos'e Ans. r. 1 — cos^ $ 24 Find the distance from (?, Fig. 48, to the centre of gravity of a spheri- cal sector generated by the revolution of a circular sector G CAj about the axis OC. Ana. k{G0+W3). 25. A circular hole with a radius r is cut from a circular disc whose radius is R\ required the centre of gravity of the remaining part, when the hole is tangent to the circumference of the disc. 26. Find the centre of gravity of the frustum of a pyramid or cone. It will be in the line which joins the centre of gravity of the upper and lower bases. Let h be the length of this line, and a and b be corresponding lines in the lower and upper bases respectively, required the distance, measured on the line A, of the centre from the lower end. a^ + ab + b* If ft = 0, we have the distance of the centre of a pyramid or cone from the base equal to ^h. 27. Find the centre of gravity of the octant of a sphere in which the density varies directly as the nth power of the distance from the centre, r being the radius of the sphere. _ n + 3 - - 2n + 8 ^ 28. Find the centre of gravity of a paraboloid of revolution of uniform density whose axis is a. An$. X — }a. 108 GENERAL PROPERTIES 77-7».> BOMB GENEEAL PROPERTIES OF THE CENTRE OF GRAVITY. 77. When a hody is at rest on a surface, a vertical through the centre of gravity will fall within the sujcyport. For, if it passes without the support, the reaction of the sur- face upward and of the weight downward form a statical couple, and rotation will result. 78. When a hody is susjpended at a point, and is at rest^ the centre of gravity will he vertically under the jpoint of sus- pension. The proof is similar to the preceding. When the preceding conditions are fulfilled the body is in equilibrium. 79. A body is in a condition of stable equilibrium when, if its position be slightly disturbed, it tends to return to its former position when the disturbing force is removed ; of unstable eguilibrium if it tends to depart further from its position of rest when the disturbing force is removed ; and of indifferent equilibrium if it remains at rest when the disturbing force is removed. Example. A paraboloid of revolution rests on a hori- zontal plane ; required the inclination of its axis. Let P be the point of contact of the para- boloid and plane, then will the vertical through P pass through the centre of gravity G, and PG will be a normal to the paraboloid. The equation of a vertical section through the centre is y^ == 2j9a?, in which x is the axis, the origin being at the vertex. Let a = AX = the altitude of the paraboloid ; 6 = GPP = the inclination of the axis ; then, AG = fa, (see example 28 on the preceding page) ; AJ!^=%a-p; hence, which will be positive and real as long as |a exceeds p. Id [80-81.] OF THE CENTRE OF GRAVITY. 109 this case the equilibrium is stable. Wlien fa exceeds j? it will also rest on the apex, but the equilibrium for this position is unstable. When ^a = j>, 6 = 90°, and the segment will rest only on the apex. When fa is less than j9, tan 6 becomes imaginary, and hence, this analysis fails to give the position of rest ; but by independent reasoning we find, as before, that it will rest on the apex, and that the equilibrium will be stable. 80. Ifi a plane material section the sum of the products found by multi- plying each elevnentary m^ass by the square of its distance from an axis, equals the sum of the similar prodaicts in reference to a parallel axis passing through the centre, plus the tnxiss mul- tiplied by the square of the distance hetween the axes. ^•^ ^s. Let AB be an axis through the centre, CD a parallel axis, D = the distance between AB and 6^Z>, d/m = an elementary mass, yi = the ordinate from AB to m, y = the ordinate from CD to m, and M = the mass of the section. Then y^={y, + Dy = y} + 2y, D + 2>». Multiply by dm and integrate, and we have ffdm =fyMm + ^Dfy^m + L^fdm. But since AB passes through the centre, the integral of y^d/m, when the whole section is included, is zero (see Eq. 715), andy^m = M\ hence, fy'dm =fy^dm -f MB". (83) Similarly, if dA be an elementary area, and A the total area, we have /fdA =fyMA + Aiy. 81. In any plane area, the sum, of the products of each ele- mentary area multiplied by the square of its distance from a/ik axis^ is least when the axis passes through the centre. 110 CENTRE OF MASS. [82.] This follows directly from the preceding equation, in which the first member is a minimum for D = 0, OENTEE OF THE MASS. 82, The centre of the mass is such a point that, if the whole mass be multijplied hy its distance from an axist it vriU eqtcal the sum of the products found hy mjultijgilying each elementary mass hy its distance from, the same axis. Let m = an elementary mass; M — the total mass ; aji, ^1, and % be the respective coordinates, of the centre of the mass, and x,y, and s the general coordinates, then, according to the definition, we have Mxi = Sm^; ) My, = 1! my 'A (84) Mhi = Hmz ; ) which being the same as (70) shows that when we consider the force of gravity as constant for all the particles of a body, the centre of the mass coincides with the centre of gravity. This is practically true for finite bodies on the surface of the earth, although the centre of gravity is actually nearer the earth than the centre of the mass is. If the origin of coordinates be at the centre of the mass, wo have i:mx = ; 2:my = ; £m3 = ; (S^) which ai^e the same as (Tla). CHAPTER IV. KON-OONOUERENT FOE0E8. 83. EQUILIBRnTM OF A SOLID BODY ACTED UPON BY ANY HDMBER OF FORCES APPLIED AT DIFFERENT POINTS AND AOXINO or DIFFERENT DIRECTIONS. ^ r/- c / B .T / ^Z)/ > '^ / A +r -X / "^1^ / *x^ y X *T, / A *X J p Fio. 64. Let A be any point of a body, at which a force F is applied, and O the oriorin of coordinates, which, being chosen arbitrarily, may be within or without tlie body. On the coordinate axes construct a parallelopipedon having one of its angles at O, and the diagonally opposite one at A. Let the typical force 7^ be in tlie first angle and acting away from the origin, so that all of its direction-cosines will be posi- tive ; then will the sign of the axial component of any force be the same as that of the trigonometrical cosine of the angle which the direction of the force makes with the axis. 112 GENERAL EQUATIONS [88.] Let a = the angle between i^and the axis of x, = (( u il u u ii ii "y, 7 = (( u u ii ii ii il "% then will the X, T, and Z-components of the force FhQ X= i^cos a, r = i^co8)8, Z — FCOS ry. The point of application of the X-component^ being at any point in its line of action, may be considered as at D^ where its action-line meets the plane yz. At E introduce two equal and opposite forces, each equal and parallel to JT, and since they will equilibrate each other, the mechanical effect of the system will be the same as before they were introduced. Combining the force + X at Z^ with — JT at E^ we have a couple whose arm is DE = y = the y-ordinate of the point A. This couple, according to Article 54, will be negative, hence, its moment is -Xy. Hence, a force + X B.t A produces the same effect upon a body as the couple — Xy, and a force + Xsit E. At the origin introduce two equal and opposite forces, each equal to X, acting along the axis of x. This will not change the mechanical effect of the system. Combining — X at O with -h X&t E, we have the couple -f Xz, and the force + X remaining at 0. Hence, a single force +X at A is equivalent to an equal jpa/rallel force at the origin of coordin- ates, and the two couples, — Xy and + Xz. . . Treating the Y-com/ponent in a similar manner, we have the force •\- Y at the origin, and tlie moments, + Yx and — Yz ; and similarly f ^r the Z-component, the force + Z at the origin, and the moments, — Zx and + 2y. But the couples + Zy SLud — Yz, have the common axis as, (b4.J OF STATICS. 113 and hence are equivalent to a single couple which is eqaal to the algebraic sum of the two ; and similarly for the others ; hence, the sioi couples may be reduced to the three following : Zy — Y2, having x for an axis Xz-Zx, « y " " " Yx-Xy, " 2 " " " hence, Jbr the single force F acting at A there may he syhstv- tuted the three axial components of the force acting at the origin of coordinates^ and three pairs of couples ha/ving for thevr axes the respective coordinate axes. If there be a system, of forces, in which Fi, Fi, i^, etc., are the forces, ^) yu ^j the coordinates of the point of application of Fi, etc., etc., etc., oi, 02) fh', etc., the angles made by i^, F^ etc., respectively with the axis of a?, A? A5 A) etc., the angles made by the forces with y, and 7ii 72) 73) etc., the corresponding angles with z ; then resolving each of the forces in the same manner as above, we have the axial components Jf = i^ cos oi + i^ cos 02 + i^ cos 03 + etc. = ^Faos a ; "j Y=Fi cos A + i^ cos /32 + i^ cos 0^ + etc. = Xi^cosyS ; V (85) Z = FiCOB yi -i- F2 cos 72 + ^ cos 73 + etc. = ^Fcoa y ; ) and the component moments Zy — Y2 = X(Fy cos 7 — i^ cos yS) = Z M Xz— Zx = X{F2 cosa — Fx cos 7) = Jf ; V (86) Yx-Xy= ^{Fx cos ^ - Fy cos a) = iT; ) in which Z, M, and iVare used for brevity. RESULTANT FORCE AND RESULTANT COUPLE. 84, Let H = the resultant of a system of forces concurring at the origin of coordinates, and having the same magnitudes and directions as those of the given forces ; 114 DISCUSSION OF EQUATIONS. [di.^ a, 5, and c = the angles which it makes with the axes aj,. y, and 3 respectively ; = the moment of the resultant couple ; d, e, and / = the angles which the axis of the resultant, couple makes with the axes x, y, and s re spectively ; ibjen JT = ^ cos a , r=i?cosJ;|- (87> Z = ^ cos £ = G coad M= Gco&e;l (88) JV= G cos/. 3 If a force and a couple, equal and opposite respectively to the resultant force and resultant couple, be introduced into the system, there will be equilibrium, and B and G will both be zero. Hence, for equilibrium, we have Z = 0; F=0; Z = 0; (89) X = 0; M=0; JV=0, (90) 85, Discussion of equations (87) and (88). 1. Suppose that the body is perfectly free to move in any^ ffvanner. a. If the forces concur and are in equilibrium, equations (87) only are necessary, and are the same as equations (60) ; hence, we will have x=o, r = o, Z=0. h Jl B = and G is finite, equations {SS) only are necessary. c. If R and G are both finite, then all of equations (87) and (88) may be necessary. 3 If one point of the body hfixed^ there can be no tra^is- lation, and equations (88) will be sufficient. 3. If an axis parallel to x is fixed in the body, there may be transjtition along tha> axis, and rotation about it : hence, the 1st of (87) and the Ist of (88) are sufficient. iw.j PROBLEMS. 115 4 If twojpoints are fixed J it cannot translate, but may rotate ; and by taking x so as to pass throiigii the two points, the equa- tion Z = is sufficient. 5. If one point only is confined to the plane xy^ the body will have every degree of freedom except moving parallel to s, and hence, all of equations (87) and (88) are necessary except the 3d of (87). 6. If three points^ not in the same straight line, are confined to \hQ plane xy, it may rotate about s, but cannot move parallel to s ; hence, the Ist and 2d of (87) and the 3d of (88) are necessary and sufficient. 7. If two aaoes parallel to x are fixed, the body can move only parallel to aj, and the 1st of (87) is suflicient. 8. If the forces are parallel to the axis of y, thei-e can be translation parallel to y only, and rotation about x and z, 9. If the forces are in the plane xy^ the equations for equi- librium become X = 2F GO& a = E Q.O& a — \\ Y= £Fcoafi= Bcoah = 0; I (91) Yx - Xy = S{Fx(i08^-Fy COS a)= 0. ) [Obs. In a mechanical sense, whatever holds a body is a force. Hence, when we say " a point is fixed," or, " an axis is fixed," it is equivalent to in- troducing an indefinitely large resisting force. Instead of finding the value of the resistance, it has, in the preceding discussion, been eliminated. When we Bay ** the body cannot translate," it is equivalent to saying that finite active forces cannot overcome an infinite resistance.] 86. Applications of equations (91). a. problems in which the tension of a 8ts/n6 18 involved. 1. A body AB, whose weight is W, rests at its lower end upon a perfectly mrvooth horizontal plans^ and at its vpper end against a perfectly smooth vertical plane : the lower end is pre- vented from sliding by a string CB, Determine the tension on the string^ and t/ie pressure upon the horisontal and vertical planes. 116 EXAMPLES. [86.3 Take the origin of coordinates at (7, the axis of x coinciding with GB^ and y with AC^ x being positive to the right, and y positive upwards. Let W = the weight of the body whose centre of gravity is at G\ B = the reaction of the vertical wall, and, since there is no friction, its direction will be perpendicular N =■ the reaction of the horizontal plane, which will be perpendicular to GB ; I = the horizontal distance from G to the vertical through the centre of gravity; t = the tension of the string. The forces may be considered positive, and the sign of the component of the force will be that of the trigonometrical function. To determine the angle between the axis of + aj and the force, conceive a line drawn from the origin of co- ordinates parallel to and in the direction of the force, then will the angle be that swept over by a line from -^-x turning left-handed to the line thus drawn. The origin of coordi- nates may be at any point, and the origin of moments at any other point. Taking the origin of coordinates, and the origin of moments both at 6> we have X =z ^ cos 0° + z5 cos 180° + W cos 270° + iTcos 90° = ; F= ^sin0° -f 2^ sin 180° + fT sin 270° + iV^sin90° = 0; Moments = - B.AG + t.O ~W.GD -{- iT. GB = 0. Keducing gives B-t = 0, -TF-fiV^=0, -^B.AG-W.GB + JV.GB = 0; B = t, [86.] EXAMPLES. 117 We see from this that the horizontal plane supports the entire weight of the piece, and that the pressure against the wall equals the tension of the string. We also notice that the forces ^ and W being equal, paral- lel and opposite, constitute a couple whose arm is DB ; and this must be in equilibrium with the couple t, CA, R ; the arm being CA, hence we have W.BB — t.CA, as before. 2. A ladder rests on a smooth horizontal plane aild against a vertical wall, the lower end being held by a horizontal string ; a person ascends the ladder, required the pressure against the wall for any position on the ladder. 3. A uniform beam, whose length is AB and weight W, is held in a horizontal position by the inclined string CD, and carries a weight P at the extremity ; required the tension of the string. Fig. 5(i Ar^-t^^.^AP-^iW). 4. A prismatic piece AB is per- mitted to turn rreely abont the lower end A, and is held by a string OE', given the position of the centre of gravity, the weight W of the piece, the inclination of the piece and string, and the point of attachment E\ required the tension of the string, and the pressure against the lower end of the beam at A. 5. A heavy jnecs AB is supported by two cords vmichpass over pulleys G and ]J, and have weights Pi and P attached to them; reqidred the inclination to the horizontal of the line AB joining the points of attachment of the cord. (Consider the pulleys as reduced to the points G and D. ) Fio. 67. 118 EXAMPLES. mi Let G, the centre of gravity of AB, be on the line join- ing the points of attachment A and B ; a = AG; h = BG; i = the angle DCM; 8 = the inclination of BA to DO; a = I)CA;siiid^ = CDB, flesolving horizontally and vertically, we have X= P, cos (180° - MCA) + P cos NVB + TTcos 270°= 0; = - Pi cos (a - *) + P cos (yg + ^) == ; (a) T= Pi sin (a - ^) + P sin (/3 + /) - W ^ 0, (b) Taking the origin of moments at 6r, we have - Fi X Gj} + P X Gj?, + W X = 0; or, ~ Pi . «^ sin (a + S) + P . ^ sin {/3 - 8) = 0. (c) The angle i is given by the conditions of the problem ; hence the three equations (a), (b), and (c) are sufficient to determine the angles a, ^, and 3, when the numerical values of the given quantities are known. The inclination will be S + i. 6. Suppose, in Fig. 58, that the strings are fastened at Osind D, and that DC\ AOy and BD are given, required the incli- nation of AB, [The solution of this problem involves an equation of the 8th degree] . 7. A heavy piece AB, Fig. 59, is free to swing about one end A, and is supported by a string BG which passes over a pulley at &, and is attached to a weight P; find the angle A CB when they are in equilibrium. 8. A weight W rests on a plane whose inclination to the horizontal is % and is held by a string whose in- clination to the plane is Q ; I'equired the relation between the tension P and the weight, and the value of the normal pressure upon the plane. Normal jpressure Fia. 60 . j^ sm ^ .^ Ans. P — ;r W; cos 6 co&{0_^ ^ cos 6 186.1 EXAMPLES. 119 h. EQtTTLIBRTUM OF PERFECTLY SMOOTH BODIES IN OONTAOT WITH EACH OTHER. 9. A heavy beam rests on two smooth inclined planes, as in Mg. 61 ; required the inclination of the beam to the horizontal^ i>nd the reactions of the respective planes. Let AC and CB be the in- clined planes; AB the beam Whose centre of gravity is at O, When it rests^ the reactions of the planes must be normal to the planes, for otherwise they would have a component parallel to the planes which would produce mo- tion. \AX,a^ = AO\ a<i—OB\ R = the reaction at A ; R= « « "i?; Tr= the weight of the beam ; a = the inclination of A C to the horizon ; ^ = « « " BO " " " ; Take the origin of coordinates at the centre of gravity G of the body, x horizontal and y vertical. The forces resolved horizontally give X == 7? cos (90° - a> + R' cos (90° + /3) + TFcos 270° = ; (a) and vertically, T = R sin (90° -a)^R' sin (90° + ^) + FF sin 270° = 0. {h) The moment of 7? sin a is, ... -f 7? sin a x ^ 6"^ sin 6. Fia. 01. u u a u u u " R' sin y9 is, . « TFcos 90°i8, . " R cos a is, . « " « i?' cos yS is, . Hence Xy — Yx — Ra^ sin a sin ^ + R'a^ sin /9 sin Q — Roi cos a cos ^ -h 7?'aa cos yS cos ^ = — Ray^ cos (a + ^) + /^'oa cos (fi-0)=O. {o) + i?' sin fi X GBBin 0. 0. — 7? cos a X AG C0& 9, + R cos j3 X GB Qjo&e 120 EXAMPLES. \^,f It is generally better to deduce the values of the moments directly from the definitions ; (see Articles 51 to 57). To da this in the present case, let fall from G the perpendiculars aG and bG upon the action-lines of the respective forces ; then bG = % sin (90° -^{/3-6))z=a2 cos (13-6); aG = «i sin (90° — {a -{- 6)) = Oi cos {a + 6); and we have the moments = — H , aG + H^ . bG = — Ra^ cos (a + ^ -h Ii'a2 cos (yS — ^) = ; as given above. Solving equations («), (J), and (c), we find sm (a + y8) ' sm (a + ff) ' . A (h cos a Qin 13 — Oi sin a cos tan ^ = 7 ; r — : ; 5 . («i 4- ^2) sm a sm p UB^E', then sin /S = sin a ; which are the conditions necessary to make the normal reaction* equal to each other. The reactions prolonged will meet the vertical through the centre of gravity at a common point i?, and if the beam be suspended at D by means of the two cords DA and DB it will retain its position when the planes AC and OB are removed. If ^ = 90°, the plane GB will be vertical, and we find ^ = TrBeca; ^' = ?^Tr = TTtan a; ' cos a tan 6 = — ^ — cot a. Oi + a^ If Oi = Oj, then sin (/3 — a) tan^ = 2 sin a sin fi [86.] EXAMPLES. . 121 I{/3= 90° and a = 0% then B' = 0, e = 90°, and R=W. A special case is that in which the beam coincides with one of the planes. The formulas do not applj to this case. 10. Two equal, smooth cy- linders rest on two smooth planes whose inclinations are a and P respectively ; required the inclination, ^, of the line joining their centres. Fio. 62. Arts. Tan 6 = ^(cot a — cot ff). 11. A heavy, uniform, smooth beam rests on one edge of a box at (7, and against the vertical side opposite ; required its inclination to the vertical. Let g be the centre of gravity. Ans. Sin^ \ Bo 12. Three equal, smooth cylinders are placed in a box, the two lower ones being tangent to the sides of the box and to each other, and the other placed above them and tangent to both ; required the pressure against the bottom and sides of the box. Fio. 64. Ans, Pressure on the hottom = total weight of the cylin- ders. Pressure on one side = ^ weight of one cylinder x tan 30°. 13. Two homogeneous, smooth, prisma- tic bare rest on a liorizontal plane, and are prevented from sliding upon it ; required their position of equilibrium when leaning against each other. Let A£ and CD be the two bars, resting against each other EXAMPLES. [86.] at B ; then will they be in equilibrium when the resultant of their pressures at B is perpendicular to the face of CD. Let5 = AB\ G = CB; x = BB: a = AD = the distance between the lower ends of the bars; W — the weight of AB ; F: = the weight of CD; jE'and G the respective centres of gravity of the bars, which will be at the middle of the pieces ; then we have 2 {a" + W-x')x' TF= g {a^ -h^ + a?) (-a^ + h^ + x^)Wi\ which is an equation of the fifth degree, and hence always ad- mits of one real root. 14. The ujpjper end of a heavy piece rests against a smooth^ vertical plane^ and the lower end. in a smooth^ spheri- cal howl / required the position of equi- librium. Let AB be the piece, BF the verti- cal surface, EA the sj^herical surface, and g the centre of gravity of the piece. ^<»- ®^* When it is in equilibrium, the reaction at the lower end will be in the direction of a normal to the surface, and hence will pasa through C, the centre of the sphere, and the reaction of the ver- tical plane will be horizontal. Let W = the weight of the piece ; r = the radius of the sphere ; a = Agxl).= Bg;l=zAB;dz:=CF\ R = the reaction of the vertical plane ; N = the reaction of the spherical surface ; i = the inclination of the beam to the horizontal ; 6 = the inclination of the radius to the liorizontaL Take the origin of coordinates at g^ x horizontal and y ver- tical ; and we have X = iV^cos ^ -t- i? cos 180° + IF cos 270° = ; T= i\rsin e -{- B sin 180° + TFsin 270° = ; A - il. 1 N/\- / , H cfM\ V y ' A ^'■. 1 Moments = + RJ) sin i — N.a^wx (^ — i) = 0; (86] INDETERMINATE PROBLEMS. 128 and the geometrical relations give, I cos i = KB = EF= d + r QO&6. From these equations, we have JV' = W Qosec 6 ; R = Trcot^; a sin {6 — i) —h cos 6 sin * = 0, which, by developing and reducing, becomes {a + h) tan i = a tan 6 ; this, combined with the fourth equation above, will determine i and 6. The position is independent of the weight of the piece, but depends upon the position of its centre of gravity. 15. A heavy prismatic bar of infinitesimal cross-section rests against the concave arc of a vertical parabola, and a pin placed at the \.r' focus ; required the position of equili])rium. Lctl=JB = length of the bar ; p = CD = one-fourth the parameter of the parabola, O being the focus, and 6 = A CD. Sr Pio. 87. Ans. 6 = 2003--^^^ 'J \jfe/ ■(fF 16. Required the form of the curve such tliat the bar will rest in all positions. Arts. The polar equation is r = i^ -f c sec 6, in which I is the length of the bar, and c an arbitrary constant. It is the equation of the conchoid of Kicomedes. C, INDETEIUIINATB PEOBLKMB. 17. To determine the jpresaures exerted hy a door ujpon its hinges. Let W = the weight of the door; a = the distance between the hinges ; b = the horizontal distance from the centre of gravity of the door to the vertical line which passes through the hinges ; ^= the vertical reaction of the upper hinge; F[ = the vertical reaction of the lower hinge ; U = the horizontal reaction of the upper hinge; Hi = the horizontal reaction of the lower hin/je ; 124 INDETEEMINATE PBOBLEMa [8&} then Xy - Yx=z Ra-Wh=: 0; which give B:= ffi = -W; and F-^ F^=W, The result, therefore, is indeterminate, but we can draw two general inferences : 1st, The horizontal pressures upon the hinges are equal to each other hut in opposite directions ; and, 2d, The "vertical reaction upon hoth hinges equals the weight of the door. It is necessary to have additional data in order to determine the actual pressure on each hinge. The ordinary imperfec- tions of workmanship will cause one to sustain more weight than the other, but as they wear they may approach an equality. The horizontal and vertical pressures being known, the^ actual pressures may be foand by the triangle of forces, II the upper end sustains the whole weight, the total pressure 1^ upon it will be — roM- h^. If each sustains one-half the weight, the pressure on each will be one-half this amount. 18. A rectangular stool rests on four legs, one being at each corner of the stool ; required the pressure on each. (The data are insufficient.) P 19. A weight P is supported by three un- equally inclined struts in one plane ; required the amount which each will sustain. Fx& 68. [Ob8. If more nnnrlitioiis are given than there aie qaantitieB to be detaN mined, they will t-iLiier be redundant or conflicting.] mi EXAMPLES. 12ft d. 8TBE8S ON FSAMBB. 20. Suppose that a triangular^tricas, Fig, 69, is loaded with A 0_®_® equal weights at the upper apices ; it is required to find the stress upon any of the pieces of the truss, [The stre»i ib the pull or piuh on a piece.] Let the truss be supported at its ends, and let I = Aa ■= ah ■=. etc., = the equal divisions of the span AB ; iV"= the number of bays in the chord AB ; L-=. Nl — AB, the span ; Pi,P2,P8, etc., be the weights on the successive apices; which we will suppose are equal to each other ; hence p=Pi =P2 = etc.; ^ = the total load ; 1^= the reaction at A ; and Fi= " " " B. 1st. T^^ere will he equilibrium among the external forces. All the forces being vertical, their horizontal components will be zero, hence X=0; r= F+ Fi - 2i> = F+ Fi ~ iV^ = ; (a) and taking the origin of moments at B, observing that the moment of the load is the total load multiplied by the horizontal distance of its centre of gravity from B, we have -Y.AB ■{- Np.\AB = ()'^ ^% V.L-Np.^L^O', .•.F=iiV^; 126 STRESS ON [86.1 which in {a) gives Y^ also equal to ^ISjp ; hence the 8U])porta sustain equal amounts, as they should, since the load is sy me- trical in reference to them, and is independent of the form of trussing. 2d. To determine the internal forces. — Conceive that th6. truss is cut hy a vertical plane and either jpart removed while we consider the remaining part. To the pieces in the plane section^ apply forces acting in such a manner as to produce tJie same strains as existed before they were severed. Consider the forces thus introduced as external^ and the prohlera is reduced to that of determining their value so that there shall he equilibrium among the new system of external forces. Let CD, Fig. 69, be a vertical section, and suppose that the right-hand part is removed. Introduce the ex- ternal forces in place of the strains, as shown in Fig. 70. FiO. TO. Let H = the compressive strain in the upper chord ; Hi = the tensile strain in the lower chord ; F = the pull in the inclined piece ; 6 = the inclination of i^to the vertical ; n = the number of the bay, bD, counting from A (which in the figure is the 3d bay) ; and D = CD = the depth of the frame. The origin of coordinates may he taken at any poiM. Take it at J , a? being horizontal and y vertical. Kesolving the forces, X^^icos 0" ^^co8 180° +F COS (270° +0+ ^cos 90°+2p cos 270° =0; r=^xsin 0° + irsin 180°+i^sin (370''+6°)H- Fsin 90°+2p 8in370°=Q; or, Hi-R-\-Fs.ine=0', (a) V-^np --Fcose = 0; (h) and the moments. - i7i'pl + RB -F.nl f JOS ^ = 0. ((?) M.J , FRAMES. 127 Eliminating i^^ between equations (h) and (c), substituting the value ofV= iNp, and reducing, give R=£^niJf-n); 2I> (d) that is, the strains on the bays of the upper chord vary as the product of the segments into which the lower chord is divided hy the joint directly under the hay considered. From (h) we have /'cos d = V- np = ^{N - 2n)j>; {e) and since 6 is constant, the stress on the inclined pieces decreases uniformly from the end to the middle. At the middle n = ^iT, and F= 0\ hence, ^br a uniform loady there is no stress on the central braces. If F were considered as a push, equation {e) would be t-egative. Eliminating II and 7^ from (a), we have For forces in a plane the conditions of statical equilibrium give only three independent equations, (a), (p) and {c) ; (or Eqs. (91) ); hence, if a plane section cuts more than three indepen- dent pieces in a frame, the stresses in that section are indeter- Tninate, unless a relation can be established among the stresses, or a portion of them be determined by other considerations. 21. If iV^= 7,i> =i?i =i?2 = etc. = 1,000 lbs., AB = 56 feet and J9 = 4 feet ; required the stress on each piece of the frame. 22. In Fig. 69, if j^i and^ are removed, and ^8 = JP4 = i>5 = 1,000 lbs., find the stress on the bay 2 — 3, and the tie 2 — b. 23. If all the joints of the lower chord are equally loaded, and no load is on the upper chord, required the stress on the w** pair of braces, counting from A, Fig. 69. Ans. i(ir — 2n -\- \)p sec 6. 128 STRESS ON A [M.1 24. A roof trvss ADB is loaded with equal weights at the equidistant joints 1, 2, 3, etc, / requi/red the stress on an/y of its members, [Obs. a load composed of equal weights on all the joints will produce the Bame stress as that of a load uniformly distributed, except that the latter would produce cross strains upon the rafters, which it is not our purpose to discuss in this place. ] Let the tie AB be divided into equal parts, Aa, ah, etc., and the joints connected as shown in the figure. The joints are assumed to be perfectly flexible. The right half of Fig. 71 may be trussed in any manner by means of ties or braces, of both, and yet not affect the analy- sis applied to the left half. Conceive a vertical section nm and the right-hand part removed. In- troduce the forces J?, H^ and F as previously explained, and the condi- tions of the problem will be repre- sented by Fig. 72. The letters of reference given below involve both figures. Let N = the number of equal divisions (bays) in AB ; n = the number of the bay ho counting from A ; I = Aa = ah, etc. ; p = the weight on any one of the joints of the rafter; V = the vertical reaction at A or B; D =• DC, the depth at the vertex; e = J2c; and* = DAG. Fia, T8. Then 1)^ = the total load; .-. F= iCiT— 1)^). \m.] ROOF TRUSS. 129 Take the origin of coordinates at A, and the origin of mo- merits at the joint marked 2. Kesolving the forces shown m Fig. 72 horizontally and vertically, we have X= J7cos(180^+ aAl)+^iCOsO°-f-i^sin(- ^2c) = 0; r= F-(n-l)i>+JJsin(180°+a^l)+^isinO°4-i^oo8(- h^o)\ = 0; or, —Ecoei -f ^i -i^sm^ = 0; ]p^_ (7i-l)^-^sin t + i^coe ^ = ; also the moments, R,b2 - V.Ab -h{n- l):p.i{n ~ 2) ^ = 0. But from Fig. 71 we have b2__ M__ {n-l)l CD'~A0^ im • Substituting in the equation of moments the value of bS found above, of F= i{J^ — l)i?, of Ah = (n -- 1) I, and reducing, give n + l)p. By means of the other two equations, and {n — 1) tan i tan ^ =1, we find B" = i{]V — n)p cosec i ; e. STRESS m A LOADED BEAM. 26. Sv^ppose that a beam is firmly fixed in a wall at one end, and that the projecting end is loaded with a weight P / required the forces in a vertical section mn^ Fig. 73. Take the origin of coordi- nates at A, X horizontal and y vertical. Take the plane sec- tion perpendicular to the axis of X. Without assuming to know the directions in which the mi.78. 130 LOADED CORD. [86.)- f trees in the section act, we may conceive them to be resolved into horizontal and vertical components. Let F be the typical horizontal force, then will hence, some of the F- forces will be positive, and the others negative. Neglecting the weight of the beam, and letting Fi be the Bum of the vertical components in nm, we have as shown in the figure. The forces, + P and — jP, constitute a couple whose arm iB Aa\ and since the F-forces are the only remaining ones, the resultant of the + F'^s and the — F^s must constitute a couple whose moment equals P.Aa with a contrary sign. [Obs. Investig-ations in regard to the distribution of the forces over the plane section belong to the Mesistance of Materials.^ f. LOADED COED. 26. Suppose that a perfectly flexible, inextensihle cord is faced at two points and loaded continuously^ according to any law ; it is required to find the equation of the curve and the tension of the cord. Assuming that equilibrium has become established, we may treat the problem as if the coi'd were rigid, by consider- ing the curve which it assumes as the locus of the point of ap- plication of the resultant. The resultant at anj^ point will be in the direction of a tangent to tiie curve at that point ; for otherwise it would have a normal component which would tend to change the form of the curve. Take the origin of coordinates at the lowest point of the curve. Let a be any point whose coordinates are x and y\ X = the sum of the x-components of all the external forceSi between the origin and a ; Y = the sum of the y-components ; [bO.J LOADED CORD. ISI t = the tension of the cord at a ; ^ = the tension at the origin. Resolving the tension {t) by multiplying it by the directiox^ coelne^ we have t --z- = the a^component of t, and t-T- = the y-component For the part Ca, equations (91) become dx ^t^^ X ^ t ds 0; Xy — Foj = 0. , [Obs. In the problems which we shall consider, the third of these eqnationa will be unnecessary, since the other two furnish all the conditions necessary for solving them.] Let aU the ajyplied forces he vertical. Then X = 0, and the first two of equations {a) become dx -A)+ t Y^t^-l^ ds 0. (5) From the first of these we have dx t -J- = to = a constant ; hence, the horizontal component of the tension will he constant throughout the length for any law of vertical loading. From the second of (i), we have 182 PARABOLIC CORD. [86.1 hence, the vertical component of the tension at any jpomt equals the total load between the lowest point and the poirU considered. 27. Let the load he uniformly dis- tributed over the horizontal. (This is approximately the condition of the ordinary suspension bridge.) Let w = the load per unit of length, then y = — wxx ^^4^ E ftnd {h) becomes ds Eliminating t gives tffly = wxdx; and integrating gives t^ = ^a» -h {C=0); ,'.a^=^y; w ^ ' w id) hence, the curve is a parabola whose axis is vertical, and whcee 2t parameter is — ^. The parameter will be constant when tQ-i- w h constant ; hence the tension at the lowest point will he the samie for all parabolas hoAivng the same pa/rameter and the samie load per unit along the horizontal, and is independent of the length of the curve. To find the tension at the lowest point, substitute in equa- tion (d) the value of the coordinates of some known point Let the coordinates of the point J. be a?i and y^ ; then (i) givei ^ = 2yi' (^ t86.] THE CATENARY. 133 To find the tension at any point we have from the first of equations {c) and the Theory of Curves , = ^^ = ^i^^ = a''i di-"^ "^ -"'• "^^ + Tojimd the tension at the highest point A, from {d) find •nbstitnte in {/), and we obtain wx. (To find to by the Theory of Moments, take the origin at A. The load on 01 will be wxi^ and its arm the horizontal distance to the centre of gpravity of the load, or \Xi ; hence, its moment wUl be \wxi^. The moment of the tension will be Uyi ; hence, toPi = itwr*! or to = ^ — , as before.) The slope (or inclination of the curve to the horizontal) may be found from equation {g) ; which gives tan * = -^\ 28. The Catenabt. A catenary is the curve assumed by a perfectly flexible string of uniform section and density, when suspended at two points not in the same vertical. Mechanically speaking the load is uniformly distributed over the arc, and hence varies directly as the arc. To find the equation, let w = the weight of the cord per unit of loijgth ; /. Y= — W8 {s being the length of the arc) ; and equations (h) become dx -t, + t^ = 0; -^+t$ = 0. ds (*) 134 THE CATENARY. 188.1 Transposing and dividing the second bj the first, gives dy w dx to and differentiating, substituting the value of ds and reducing, give W , \dxj . - dx = I sag* Integrating gives w to or, passing to exponentials, gives ..-^.[|.v/..g]; ;ive8 ^ ^ da? dx dx or, 1 + from which we find d d which integrated gives U) which is the equation of the Catenary. Eliminating -~ between equations (^) and (J), we find ds dx (86. J THE CATENARY. 135 the integral of which is (m) which gives the length of the curve. The following equations may also be found X = - lOga h \/ 1 + —TT- V I ~ w dx If ^ — the inclinatiou of the curve to the vertical, theo a; = s tan Q log^ cot \Q, The tensions, t and 4, are so involved that they can be de- termined only by a series of approximations. The full devel- opment of these equations for practical purposes belongs to Applied Mechanics. The catenary possesses many interesting geometrical and mechanical prop- erties, among which we mention the following : — The centre of gravity of the catenary is lower than for any other cnrve of the same length joining two fixed points. If a common parabola be rolled along a straight line, the locus of the focus will be a catenary. According to Eq. (k) it appears that if the origin of coordinates be taken directly below the vertex at a distance equal to t^ -^ w, the constant of integra- tion will be zero. (This distance equals such a length of the cord forming the catenary as that its weight will eqnnl the tension at the lowest point of the curve). A horizontal line through this point is the directrix of the cateruiry. The radius of curvature at any point of the catenary equals the normal at that point, limited by the directrix. The tension at any point equals the weight of the cord forming the cate- nary whose length equals the ordinate of the point from the directrix. If an indefinite number of strings (without weight) be suspended from a catenary and terminated by a horizontal line, and the catenary be then drawn out to a straight line, the lower ends of the vertical lines will be in the arc of a parabola. If the weight of the cord varies continuously according to any known law the curve is called Catenarian, 186 LAW OF LOADING. my 29, To detenmne the equation of the Catena/ricm cttrve of uniform density in which the section varies directly as th4 tension. Let ^ = the variable section ; 5 = the weight of a unit of volume of the cord ; G = the ratio of the section to the tension ; then T=-/BMs; k = ct; .\T= -Bc/tds; which substituted in (b) and reduced, gives Bey = loge sec c8x, for the required equation. g, LAW OF LOADING. 30. It is required to find the law of loading so that the action-line of the resultant of the forces at any jpoint shall he tangent to a given curve. Assume the loading to be of uniform density, and the variations in the load- ing to be due to a variable depth. In Fig. 76, let O be the origin of coordi- nates ; Z = a5 = the depth of loading over a point whose abscissa is a? ; d=^ the depth of the loading over the ri- gin, and 3 = the weight per unit of volume of the loading, then ]r= -fhZdx', which in Eq. {b) gives Transposing, and dividing the latter by the former, gives ^ = ^ fZcb>; dx uJ ' [86.] LAW OF LOADINa 187 which, difFerentdated, gives da? U Bat, from the Theory of Onrves, we haye d?y _ V d^J _ sec'i dx^ "" p ~~ p in which p is the radius of curvature, and i is the angle between a tangent to the curve and the axis of x. From these we read- ily find „ to sec^^ 8 p At the origin p z=z pQ, i = 0, and Z =: d; which values sub- stituted in the preceding equation give ^ „ J secFt , ^ /. Z = dpQ . (n) Discussion. For all curves which have a vertical tangent, we have at those points i = 90** ; .*. sec t = 00 , and, if p is finite Z= oo; hence, it is practically impossible to load such a curve through- out its entire length in such a manner that the resultant shall be in the direction of the tangent to the curve. A portion of the curve, however, may be made to fulfil the required con- dition. Let the given cv/rve he the arc of a circle / then p = p^, and equation {n) becomes Z = d sec' i, from which the upper limit of the loading may be found. For 138 EXAMPLES OF A [86. J Bmall angles secH will not greatly exceed imitj and hence, the upper limit of the load will be nearly parallel to the arc of the circle for a short distance each side of the highest point. At the extremities of the semicircle, i = 90°, and Z = ao , If the given curve he a parahola^ we find Z = ^, that is, the depth of loading will be constant ; or, in other words, uni- formly distributed over the horizontal. This is the reverse of Prob.*27. (The principles of this topic may be used in the construction and loading of arches.) 31. Let the tension of the cord he uniform. We observe in this case that the loading must act normally to the curve at every point, for if it were inclined to it, the tangential component would increase or decrease the tension. Let j!? = the normal pressure per unit of length of the arc; then pds = the pressure on an element of length, and this mul- tiplied by the direction-cosine which it makes with the axis of Xy and the expression integrated, give I j)ds (y J = / j)dx = the aj-component, and / pdy = the y-component of the pressures, hence, equations (a), p. 131, become dx — 4 + J pdx + t -T— = ; differentiating which, give dx T86.J NORMALLY PRESSED ABO. 139 Transposing, squaring, adding and extracting the square root, give that is, the normal pressure varies inversely as the radius of cwrvature, 1. If a string be stretched upon a perfectly smooth curved surface by pulling upon its two ends the nortnal pressure upon the surface will vary inversely as the radins of curvature of the surface, the curvature being taken in the plane of the string at that point. 2. If /3 be constant p will be constant ; heiice, if a circular cylinder be immersed in a fluid, its axis being vertical, the nor- mal pressure on a horizontal arc being uniform throughout its circumference, the compression in the arc will also be constant. A. THE LAW OF LOADING ON A NORMALLY PBESSED ARC BEING GIVEN, REQUIRED THE EQUATION OF THE ARC 32. The ties of a suspension . ^ bridge being normal to the cu?^e of the cable, and the load uni- form along the span, required b * * *~~e' the equation of the curve of fio. 77. the cdblt Ana, <1 -h (i*v/i)'-[Hi*v(iH}S^ the origin being at O, x horizontal and y vertical. If tan i = -p, and po = the radius of curvature at the vertex, then X = ^pq (1 + cos' i) sin i, y == kpQ sin' ico&i, (See solution by Prof. S. W. Robinson, Journal of the Franklin Institute, 1863, voL 46, p. 145 ; and its application to bridges and arches, voL 47, p. 162 and p. 361.) 140 NORMALLY PRESSED ARC. [86 i 33. A perfectly flexible, inextensible trough of mdefmit6 length is filled with ajhiid, the edges of the trough heing pa/T' aUel and sujpjported in a horizontal jplane ; required the equa, tion of a cross section. The length is assumed to be indefinitely long, so as to elimin- ate the effect of the end pieces. The pressure of a fluid against a surface is always normal to the surface, and varies directly with the depth of the fluid. The actual pressure equals the weight of a prism of water whose base equals the surface pressed, and whose height equals the depth of the centre of gravity of the said surface below the surface of the fluid. The problem may therefore be stated as follows ; — Required the equation of the curve assumed hy a cord fixed at two points in the same hori- zontal, and pressed normally hy forces which vary as the verti- cal distance of the point of application helow the said hori- zontal. Let A and B be the fixed points. Take the origin of the coordinates at D, midway be- tween A and B, and y positive downwards. Let 8 be the weight of a unit of volume ; then p = By, which in equation (o) gives t = Syp, and for the lowest point t = BBpo ; in which D is the depth of the lowest point and po the radius of curvature at that point ; .-. gyp = Si?p„, or ^^==1. But from the Theory of Curves we have p~\ ^dx") d^' which substituted above, and both sides multiplied hy dy^ may be put under the form {8a.l HYDBOSTATIG TBOUOH. 141 the integral of which is But ^ = 0, for y = 2) ; .: O = ^^~ ; which gabstitnted and the eqaation reduced gives ^ Squaring and reducing, gives These may be integrated by means of Elliptic Fhmctiona. Making y = D cob «^, and g = ^—y they may be reduced to known forms. Using Legend/r^s notation, we have (See Article by the Author in the Jowmal of the FrowMMk Institute, 1864, vol. 47, p. 289.) CHAPTER V. RELATION BETWEEN THE INTENSITIES OF FORCES ON DIFFEEKHT PLANES WHICH CUT AN ELEMENT. 87. Distributed Forces are those whose points of applica- tion are distributed over a surface or throughout a mass. The attraction of one mass for another is an example of the latter, some of the properties of which have been discussed in the Chapter on Parallel Forces ; similarly, when one part of a body is subjected to a pull or push, the forces are transmitted through the body to some other part, and are there resisted by other forces. If the body be intersected by a plane, the forcea which pass through it will be distributed over its surface. Plaiies having different inclinations heing passed through an element^ it is jprojposed to find the relation hetween the inte7isi^ ties of the forces on the different planes. 88, Definitions. Stresses are forces distributed over a sur- face. In the previous chapters we have assumed that forces are applied at points, but in practice they are always distrib- uted: A strain is the distortion of a body caused by a stress. Stresses tend to change the form or the dimensions of a body. Thus, 2i.j)ull elongates, ^push compresses, a twist produces tor- sion, etc. (See llesistance of Materials^ A simple stress is a pull or thrust. Stresses may be com- pound, as a combination of a twist and a pull. A direct simple stress is a pull or thrust which is Dormal to the plane on which it acts. A pull is Qen^idiCYQdi positive^ and a push, negative. The intensity of a stress is the force on a unit of area, if it be constant ; but, if it be variable, it is the ratio of the stress on an elementary area to the area. To form a clear conception of the forces to wliich an element 18 subjected, conceive it to be removed from the body and then r89,90.J RESOLVED STRESSES. 143 subjected to such forces as will produce the same strain that it had while in the body. 89, Formulas for the iNTENsrfT of a stress. Let 7^ be a direct simple stress acting on a surface whose area is A, and jp the measure of the intensity, then F wlien the stress is uniform, and (92) jp = -T-j, when it is variable. If the stress be variable we will assume that the section is so small that the stress may be considered uniform over its sur- face. 90. Direct stress resolved. Let the prismatic element ABj Fig. 79, be cut by an oblique plane DK Let the stress F be simple and direct on the surface OB, and ^ = the normal component of i^^on T = the component of F along the plane J)E, which is called the tangential component ; = FON — the angle between the action -line of the force and a normal to the plane DE, and is called the obliquity of the jplane ; Fxa. m A = the area of CB, and A' that of DK Then, according to equations (62), we have iV^= Ffto^e-y T = Fsin d. From the figure we have A' r=z A sec ^, hence, on the plane DE, we have 144 SHEARING STRESS [M.] JVormal mUrmtt/,jpn= --T} = -5 >, = p cos' ^; .^ ^ S6C U Tangential intensity, jpt =-jj= -. ^ z=.jp siu d cos ^. (93) '""■^'1 (95) Pass another plane perpendicular to DE, having an obli- quity of 90° — Q ; then, accenting the letters, we have y„=i?8in2^; ) y, =i>cos(98in^. ) ^^^ This result is the same as if a direct stress acting upon a plane perpendicular to GB, having an obliquity of 90° — ^ in reference to DE, be resolved normally and tangentially to the latter. Combining equations (93) and (94) we readily iind that is, when an element {or hody) under a di/rect simple stress is intersected hy two jplanes the sum of whose obliquities is 90 degrees, the sum of the intensities of the normal comjponents of the stress equals the intensity of the direct sim/ple stress, and the intensities of the tangential stresses are equal to each other, 91. Sheaimng stress. The tangential stress is commonly called a shearing stress. It tends to draw a body sidewise along its plane of action, or along another plane parallel to its plane of action. Its action may be illustrated as follows : — Suppose that a pile composed of* thin sheets or horizontal layers of paper, boards, iron, slate, or other substance, having friction between the several layers, be acted upon by a horizontal force applied at the top of the pile, tending to move it sidewise. It will tend to draw each layer upon the one immediately beneath it, and the total force exerted between each layer will equal the applied force, and the resistance at the bottom of the pile will be equal and opposite to that of the applied force. If other horizontal forces are applied at different points along the ver- tical face of the pile, the total tangential force at the base of the pile will equal the algebraic sum of all the applied forces. A shearing stress and the resisting force constitute a cowpUy T»l. SHEARING STRESS. 145 and as a single couple cannot exist alone, so a pair of shearing stresses necessitate another pair for equilibriuni When the direct simjple stresses on the faces of a rectangular jparallelopvpedon are of equal intensity, the shearing stresses will he of equal intensity. Let Fig. 80 represent a paralellopipedon with direct and shearing stresses applied to its several faces. At pre- sent suppose that all the forces are parallel to the plane of one of the faces, as a^fe, and call it a jplane of the forces ; then will the 'planes of action, which, in this case, will be four of the faces of the parallelopipe- don, be perpendicular to a plane of the forces. If the direct stress + F = — i^ and + i^' = — F' , they will equilibrate each other. The moment of the tangential force T, will be ©< X areafc x ah ; and of T p\ X area ac x hf The couple T.ah tends to turn the element to the right and T'.hf to the left, hence, for equilibrium, we have ;Pt X areafc x ah = p't x area ac x hf; but area fcxah=: area ac x hf =^ the volume of the ele- ment, hence The effect of a jpair of sheaHng stresses is to distort the element, changing a rectangular one into, a rhomhoid, as shown in Fig. 81. Direct stresses are directly opposed to each other in the same plane or on opposite surfaces ; shearing stresses act or. parallel planes not coincident 10 (96) \ ^ Via. 8L 146 NOTATION. [92, 93.) 92. Notation. A very good notation was devised by Co riolis, which has been used since 1837, and is now commonly employed for the general investigations on this subject. It is as follows: — Let J? be a typical letter to denote the intensity of a stress of some kind ; p^ the intensity of a stress on a plane normal to V ; ^aa? the intensity of a stress on a plane normal to x and in a direction parallel to a?, and hence indicates the intensity of a direct simple stress ; and p^^ the intensity of a stress on a plane normal to x but in the direction of y, and hence indi- cates the intensity of a shearing stress. Or, generally, the first sub-letter indicates a normal to the 'plane of action and the second one the direction of action. Hence we have INTENSITEES OF THE FOE0E8 parallel to X y Vw JPzy PvA on a plane normal to - X z. li direct stresses only are considered, one sub-letter is suffi* cient; && p^, py, or p^. 93. Tangential stress resolved. Let T be the tangential stress on the right sec- tion AB = A, the section being normal to y, then p^= T-^ A. Let CD be an oblique section, normal to tte axis y' ; x' and x being in the plane of the axes y and y' ; then will the angle be- tween y and y' be the obliquity of^ the plane OB. This we will denote by (yy'). Let the tangential force be parallel to the axis of x. Resolving this force, we have Normal component on CD = T sin (yy') ; Tangential component on CD — T cos ^\ Fig. 82. I»l] RESOLVED STRESSES. 147 Dividing each of these by aj^ea CD = AB -r- coe (^^ we have Narmai interudty = py'y' = ^^jj^^T^^^ =Pyx^ {fftf) cos [yy'){ AB J/ cos^ ivu' ) Tangential intensity =pyt^t = -/ = pyx cos' {yy')\ a/tea An (97) and for a tangential stress on a plane normal to a?, resolved upon the same oblique plane GD^ we have JP'v'y' = I^^ cos (yy') sin {yy') ; = 7)^. snr Pl/^ -JPxu W)' (98) If the tangential stresses on both planes (one normal to y, and the other normal to x) are alike, and the obliquity of the plane CD less than 90°, the resultant of their tangential components will be the difference of the two components, as given by equations (97) and (98); that is, it will be ^y/g/ ""i^Va/; bat the normal intensity will be the sum of the components as given by the same equations. The reverse will be true in re- gard to the direct stresses. 94. Let a body he svhjected to a direct si/mple stress; it ts required to Jind the stresses on any two jplanes perpendicular to one another and to the plane of the forces / also the intensity of the stress on a third plane perpendicular to thi plane of the forces ; and the normal and tangential cornpo7t£nts on that plane. Let the forces be parallel to the plane of the paper ; A O and OB, planes perpendicular to one another and to the plane of the paper, having any obliquity with the forces. Let the axis of x coincide with OB, and y with AG. Let AB be a third plane, also pei-pendicular to the plane of the paper, cutting the other Pio. 88. planes at any angle. Take y' perpendicular to AB and af* parallel to it and to the plane of the paper. 148 RESOLVED STRESSES. I»5. The oblique forces may be resolved normally and tangen- tially to the planes AO and OB, by means of equations (93) and (94). The problem will then be changed to that shown in Fig. 84, in which .one set of stresses is simple and direct, and the other set tangential; and, according to Article 91, the intensity of the shearing stress on the two planes will be the Fza. 84. same ; hence, for this case The intensity of the total normal stress on the plane AB will be the sum of the normal components given by equations (93), (94), (97) and (98), and the total tangential stress will be the sum of the components of the tangential stress given by the same equations ; hence V yy = Pxx sin'' (pp') + pyy cos' {yy') + 2pxy sin W) cos {yy') ; ) .gg. Vy'x -\Pxx~Pyy\ sin {yy) cos {yy') + Pxy { cos'' {yy') — sin» {yy'^ \ • S The resultant stress on ^^ will be, according to equation (46), e being 90°, and the inclination of the resultant stress to the normal, y\ will be tan(^')=J^; (101) 95. DlSOTTSSION OF EQUATIONS (99). A. Mnd the inclination of the j^lane on which there is no tangential stress. In the 2d of equations (99) make^^a-- = 0, and representing this particular angle by (yy'), we find tan 2(yy') = _ 2sin {yy') cos {yy') __ 2/?, cps^ (2/2/') - sin^ {yy') _P'— , (102) rW.] PRINCIPAL STRESSES. l<k» which gives two angles differing from each other by 90°, or, the planes will be perpendicular to one another. Hence, in every case of a direct simple stress upoii a pair of planes perpendicular to one another and to a plane of ths stresses, there are two planes, also perpendicular to one another and to the plane of the stresses, on which there is no tangential stress. These two directions are QdXlQdi princtpaZ axes of stress. Principal axes of stress are the normals to two planes per- pendicular to one another on which there is no tangential stress. Principal stresses are such as are parallel to the principal axes of stress. (In some cases there is a third principal stress perpendicular to the plane of the other two.) The formulas for the stresses become most simple by refer- ring them directly to the principal axes. a. Let one of the direct stresses he zero^ Equation 102 gives tan 2(ty') =t ^ (103) h. Let one of the direct stresses he a puUy <md the oth&r a push. Then tan 2(YY') = -^^ (104) e. Let them act in opposite senses and equal to each other. Then tan2(YT')=:-^. (105) d. Let there he no tangential stress on the original pla/M$^ OTp^=0. Then, tan 2(yy') = ; /. (ty') = or 90* ; and the original planes are pr^incipal planes. 150 .RESOLVED STRESSES. [1W.J e. Let there he no direct stresses. Then, tan 2(yy') = oo ; or (yy') = 45° or 135° ; (106) that is, if on twoj>lanes, jperjpendiGular to one another' and to iht plane of the stresses, there are no direct stresses, then will the stress on two jplanes, perj^endicular to one another and to the plane of the stresses, whose inclination with the original planes is 45°, he simple and direct. f. Let the direct stresses he equal to one another and act in the same sense^ and let there he no shearing on the original planes. Then tan2(YY')=^; and (yy') is indeterminate; hence, in this case every plane perpendicular to a plane of the stress will be a principal plane Examples. - 1. A rough cube, whose weight is 550 pounds, rests on a horizontal plane. A stress of 150 pounds applied at the upper face pulls vertically upward, and another direct stress of 125 pounds, applied at one of the lateral faces, tends to draw it to the right, while another direct stress of 50 pounds tends to draw it to the left ; required the position of the planes on which there are no tangential stresses. If the cube is of finite size it will be necessary to modify the problem, in order to make it agree with the hypothesis under which the formulas have been established. The force of gravity being distributed throughout the mass, would cause a variable stress, and the surface of no shear would be curved instead of plane. We will therefore assume that the cuhe is without weight, and the 550 pounds is applied directly to the lower surface. Then the vertical stress will be 150 pounds, the remaining 400 pounds being resisted directly by the plane on which it rests, and so far as tlie present problem is concerned, only produces friction for resisting the shearing stress. The direct horizontal ^95 1 EXAMPLES. 151 stress will be 60 pounds, the remaining 76 pounds producing a shearing on the horizontal plane. The former force tends to turn the cube right-handed by rotating it ^bout the lower right- hand comer, thus producing a "'^action or vertical tangential stress of 75 pounds. Let the area of each face of the cube be unity, then we have Pxy = ^^ pounds ; pg^^ = 50 pounds ; jpyy = 150 pounds ; and these in (102) give tan2(T.') = gJ^g5 = 1.5; .-. (ttO = 4- 28° 9' 18'', or - 61° 60' 42". If the body be divided along either of these planes, the forces will tend to lift one part directly from the other without producing sliding upon the plane of division. 2. A rough body, whose weight is 100 pounds, rests on an inclined plane ; required the normal and tangential components on the plane. (Use Eq. (93).) 3. A block without weight is secured to a horizontal plane and thi-ust downward by a stress whose intensity is 150 pounds, and pulled towards the right by a stress whose intensity is 150 pounds, and to the left with an intensity of 100 pounds ; re- quired the plane of no shear. 4r. A cube rests on a horizontal plane, and one of its vertical faces is forced against a vertical plane by a stress of 200 pounds applied at the opposite face, and on one of the other vertical faces is a direct pulling stress of 75 pounds, which is directly opposed by a stress of 50 pounds on the opposite vertical face ; required the position of the plane of no shear. In this case the weight of the cube would be a third princi- pal stress, but it is eliminated by the cx)nditions of the problem. The shearing stress is 25 pounds; and because the direct stresses are unlike, we use Eq. (104). 5. A rectangular parallelopipedon stands on a horizontal p.ane, and on the opposite pairs of vertical faces tangential 162 PLANES OF [W.y Btresses of equal intensities are applied ; required the position of the plane of no shear. (See Eq. (106).) 6. In the preceding problem find the intensity of the direct stress on the plane of no shear. (Substitute the proper quanti- ties in the 1st of (99).) B. To find the planes of action for maximum and minimum twrmal stresses, and the values of the stresses. Equate to zero the first differential coefiicient of the 1st of Equations (99), and we have 2p^ cos (2/2/0 sin (2/2/') - 2^^ sin (2/y') cos (yy') ) [ (107) - 2p^ sin2 (2/2/') + 2p^ cos2 (yy') = ; J /. tan 2(^2/') - ""^-^^ ' Paex jPyy which, being the same as (102), shows that on those planer which have no shearing stress, the direct stress will be either a maximum or a minimum. Testing this value by the second differential coefficient, we find that one of the values of (yy') gives a maximum and the other a minimum. Comparing (107) with the second of (99), shows that the jurat differential coefficient of the value of the direct stress on ant^ plane is twice the shearing stress on that plane. From (107), observing that cos (2/y') = 4/I — sin^ (yy')) we find sin* (ytO = i- I 1 =F /==^^=^|^== i ; (108) and tliese values in the 1st of (99), and the maximum and minimum values designated by p^', give in which the upper sign gives the maximum, and the lower the minimum stress. These are jprincvpal stresses^ and we denote them by one sub-letter. [95.] MAXIMUM STRESS. 158 d Tff^ — 0, we have {yy') = 0° or 90°, a8 we should. ft. Ifjpyy = 0, we have (110) maximum, p^, = ip^ + Vip'^xx + l^xy \ \ minimum, jp^ = ip^ - ^/ip^^-^-j^^; J hence, the maximum normal stress will be of the same kind as the principal direct stress, j^aa,; that is, if the latter is &puUy the former will also be a pull, and the minimum principal stress will be of the opposite kind. c. If there are no direct stresses p^a will also he zero, and we have (TT') = 45°orl35°; and m^aximum^p^t =Pxy = —Pr' for minimum ; that is, the principal stresses will have the same intensity aa the shearing stresses, and act on planes perpendicular to one another, and inclined 45° to the original planes. EXAMPLBS. 1. Suppose that a rectangular box rests on one end, and that one pair of opposite vertical sides press upon the contents of the box with an intensity of 20 pounds, and the other pair of vertical faces press with an intensity of 40 pounds, and that horizontal tangential stresses, whose intensities are 10 pounds, are applied to the vertical faces, one pair tending to turn it to the right, and the other to the left ; required the position of the vertical planes of no shearing, and the maximum and minimum values of the direct stresses. 2. For an application of Equations (103) and (110) to the stresses in a boam, see the Author's Resistance of Materials, 2d edition, pp. 236-240. O. To find the position of the plcmes of maxvmum and tninvfnum shearing. 164 PRINCIPAL STRESSES. [95. | Equate to zero the first differential coefRcient of the second of (99) and reduce, denoting the angles sought by {YY'), and we find, - cot2(ZJr') = tan2(YT'); .-. 2Fr' = 2(YY') + 90°; or, YY' = yt' + 45° : that ts, the jplanes of maximum and minimum ahea/r make angles of AiO degrees with the pklncipal planes. D, Let the planes he pkincipal sections. Then the stresses will be principal stresses^ and ^a^ = 0. Using a single subscript for the direct stresses, equations (99) become py' = p^ sin2 {yy') + p^ cos^ {yy') ; j, ^ Pv'x = {Px - Pv) sin W) cos {yyy ) a. Let^as =Pvf *^^^ I>v'=Px; and^^^ = 0; that is, when two principal stresses are aliJce and equal on a pair of planes perpendicular to the plane of the stresses, the normal intensity on every plane perpendicular to the plane of the stresses will he egual to that on the principal planes, and there will he no shearing on any plane. This condition is realized in a perfect fluid, and hence very nearly so in gases and liquids, since they offer only a very slight resistance to a tangential stress. If a vessel of any liquid be intersected by two vertical planes perpendicular to one another, the pressure per square inch will be the same on both, and will be normal to the planes ; hence, according to the above, it will be the same upon all planes traversing the same point. This is only another way of stating the fact that fluids press equally m all directions. h. To find the planes on which there will he no normal pre^- 8ure, For this^y^ in (111) will be zero ; (Ml PBOBLEM. 155 .•.tan(yy')=\/f l/^^ Jrv which, being imaginary, Bhows that it is impossible when the stresses are alike ; but if they are unlike^ we have tan(vy')='\/^V^= \/5- If j>, = "Pyy then iyy') = 45°, and the 2d of (111) gives which shows that when the direct stresses are unlike and of equal intensity on planes perpendicular to one another, the shearing stress on a plane cutting both the others at an angle of 45 degrees, will bo of the same intensity. Let (yy') = 45°, or ISS**, then (111) become in the latter of which the upper sign gives a maximum^ and the lower a minimum value. Using the upper sign, wo find Px = Pv 96. Problem. Find the plane on which the obliquity of the stress is greatest^ the intensity of that stress^ and the angle of its obliquity. Let the stresses be principal stresses and of the same kind, and <i> the angle of obliquity of the required plane to the stress ; then ■in<^ ^ i^«""i^y . (he intensity = ViP'Pv)^ ^^^ t^© angle be- Px+Pv tween the principal plane x and the required plane = 45®— J^ 156 CONJUGATE STRESSES. [97.^ If the principal stresses are unlike, then sin (f)' =^ — ^; the mtensity = ^ —jpxJPv^ and the angle be- tween the principal plane a?, and the oblique plane = 45° — ^^V Example. If a body of sand is retained by a vertical wall and the mtensity of the horizontal push is 25 pounds, and of the ver- tical pressure is 75 pounds ; required the plane on which the resultant has the greatest obliquity, and the intensity of the stress on that plane. CONJUGATE STRESSES. 97. A pair of stresses, each acting parallel to the plane of action of the other, and whose action-lines are parallel to a plane which is perpendicular to the line of intersection of the planes of action, are called conjugate stresses. Thus, in Fig. 85, one set of stresses acts on the plane YY^ parallel to the plane XX^ and the other set on XX, parallel to YY. In a rigid body the intensities of these sets of stresses are independent of each other; for each set equilibrates itself. Princijpal stresses are also conjugate. There no^ay be three conjugate stresses in -a body, and 07ily three. For, in Fig. 85, there may be a third stress on the plane of the paper, which may be parallel to the line of intersection of the planes XX and YY, and each stress will be parallel to the plane of the other two. A fourth stress cannot be introduced which will be conjugate to the other three. Conjugate stresses may be resolved into normal and tangen- tial components on their planes of action, and treated accordhig to the preceding articles. The fact that the stresses have the same obliquity, being the complement of the angle made by the planes, simplifies some of the more generul problems of m-] GENERAL PROBLEM. 167 GENERAL PROBLEM. 98. Given the stresses on the three rectangular coordmats jplaries ; required the stresses on any obligiie plane in any re- quired direction. As before, the element is supposed to be indefinitely smalL Let ahc be the oblique plane, the normal to which designate by n. The projection of a unit of area of this plane on each of the coordinate plains, gives respectively cos {nx\ cos {ny\ cos {m). The direct stress parallel to x acting on the area cos {nx) will give a stress of jpxx cos {nx\ and fio. 86. the tangential stress normal to y and parallel to x will produce a stress j?j^ cos {ny), and similarly the tangential stress normal to 2 and parallel to x gives ^« cos {m) ; hence the total stress on the unit normal to n and parallel to x will be Pnx = i>asB COS {nx) + py^ COS {ny) + jp«j cos inz) ; ' similarly, jpny = J>xv cos {nx) ^-Jpyy COS {ny) + p^ cos {nz) ; Pnt = p„ cos {i-vx) -^pyt cos i^y) + j9aj cos {nz). Let these be resolved in any arbitrary direction parallel to s. To do this multiply the first of the preceding equations by cos («»), the second by cos {8y\ and the third by cos (sz)^ and add the results. For the purpose of abridging the formulas, let cos {nx) be written C/iaj, and similarly for the others. Then we have Pn» =pxxOnxGsx + pyyOnyOsy -\- p^CnzOsz \ -\-pyz (CnyCsz -h GnzCsy) -{- p,x{CnzGsx v (115) -h CnxCsz) + pxy (CnxCsy + CnyCsx). ) This expression being typical, we substitute x' for n and «, and thus obtain an expression for the intensity on a sui face normal to x' and parallel to x'. Or generally, substitute suc- cessively a?', y\ z' for n and «, and we obtain the following formulas : \ (11*) 168 GENERAL FORMULAS. [98.1 DIBEOT STRESSES. + ^^zxOzx'Qxx'-\- '^iPseyOxx'Qyx ; + 2/?««Csy' Ca;2/'+ ^jp^Cxy'Gyy' ; + 2/?aa.C2;3'Ca73' + 2j>xy^xz'Gyz' ; lANQENTIAL STRESSES. j?^^ =^a^Ca?y'Caj3' -^ jpyyOyy'Oyz' + p^Csy'Gzs' + Czz'Gxy') + p:^(CxyVyz' + Cajs'CyyO ; ^^5B, = ^^aaCic^'CiCic' + jpyyQyz'Qyx' + ^^g* C^s'Csaj' + i?y.(Cys'C^a5' + Qzz'Qy^)^p,JS^zz'Qxx'-\-QzxlQxz') +J?xy {Cxz'Qyvf + Cxx'Gyz') ; j>j^y, = j?a«Ca3a;'Cajy' + jpyyOyx'Oyy' + j^Jdzx'Qzy^ +2>ylCyxVzy'+Oyy'Czxf)+:p,JS^zx'(:^'+Qzy'Q^^ + j[>xy{pxx'Cyy' + Ca^'Cyaj'). It may be shown that for every state of stress in a hody there are three planes perpendicular to each other ^ on which the stress is enti/rely nomuiU [These equations are useful in discussing the general Theory of the Elasticity of Bodies,'] These formulas apply to oblique axes as well as rights only it should be observed when they are oblique that J9yv is not a stress on a plane normal to y\ parallel to z\ but on a plane nor- mal to x' resolved in the proper direction. CHAPTER VI YIBT17AL VELOCITIES. 99, Bef. If the point of application of a force be moved in the most arbitrary manner an indefinitely small amount, the projection of the path thus described on the oriirinal action-line of the force is called a h virtual velocity. The product of the force / I ^ into the virtual velocity is called the virtual ^^ ^^ moment. Thus, in Fig. 87, if a be the point of application of the force F, and ah the arbitrary displacement, ac will be the virtual velocity, and F.ao the virtual moment. The path of the displacement must be so short that it may be considered a straight line ; but in some cases its length may be finite. If the projection falls upon the action-line, as in Fig. 87, the virtual velocity will be considered positive, but if on the line prolonged, it will be negative. 100. Prop. Jf several concurring forces are in equilibnum^ the algebraic sum, of their virtual moments will he zero. Using the notation of Article (47), and in addition thereto let I be the length of the displacement ; and p, q^ and r the angles which it makes with the respective coordinate axes; then will the projections of I on the axes be I cos^, Z cos 2', I cos r, respectively. Multiplying equations (50), by these respect- ively, we have F^QO&a^l cos^ + F2C0& Oil cos^ + etc. = ; 2^ cos A ^ cos ^ + i^ cos ^^IcoAq r etc. = ; F QO& fill cos /• 4- jpj cos 72 ^ cos r + etc. = 0. 160 VIRTUAL VELOCITIES. [101.] Adding these together term by term, observing that cos a cos^ -h cos fi coa q -\- cos 7 cos r = cos (i^Q ; which is the cosine of the angle between the action-line of JF* and the line I; and that I cos (I^l) = Bf (read, variation J^ is the virtual velocity of i^ we have F,df, + 7^8/2 + i^S/3 + etc. = ^F8f = ; (116) which was to be proved. 101. ^f Cbi^y number of forces in a system are in equilHh rium, the sum of their virtual moments will he zero. Conceive that the point of application of each force is con- nected with all the others by rigid right lines, so that the action of all the forces will be the same as in the actual problem. If any of the lines thus introduced are not subjected to stress, they do not form an essential part of the system and may be cancelled at first, or considered as not having been introduced. Let the system receive a displacement of the most arbitrary kind. At each ^^**' ^' point of application of a force or forces, the stresses in the rigid lines which meet at that point, combined with the applied force or forces at the same point, are necessarily in equilibrium, and by separating it from the rest of the system, we have a system of concurrent forces. Plence, for the point B^ for in- stance, we have, according to (116), Flf + F^hf + etc. + BCBBG+BABBA + BDhBD = ; in which BO, etc., are used for the tension or compression which may exist in the line. But when the point O is consid- ered, we will have BC8BC with, a contrary sign from that in the T)receding expression, and hence their sum will be zero. Proceeding in this way, as many equations may be estab- lished as there are points of application of the forces ; and adding the equations together, observing that all the expressions which represent stresses on the lines disappear, we finally have 2FBf=0. - (117) {101.1 EXAMPLES. 161 The converse is evidently true, that when the sum of the vir* iual inoments is zero the system is in equilibrium. Equations (116) and (117) are no more than the vanishing equations for work. If a system of forces is in equilibrium it does no work. This principle is easily extended to Dynamics. For, the work which is stored in a moving body equals that done by the impelling force above that which it constantly does in overcoming resistances. Thus, when friction is overcome, the impelling forces accomplish work in overcoming this resist- ance, and all above that is stored in the moving mass. Letting R be the resultant of all the impressed forces producing motion, and s the path described by the body, we have Rhr^ Sm-^8s = 0. (118) This is the most general principle of Mechanics, and M. Lagrange made it the fundamental principle of his celebrated work on Mecanique Analytique, which consisted chiefly of a discussion of equation (118). Examples. 1. Determine the conditions of equilibrium of the straight lever. Let AB be the lever, having a weight P at one end and W at the other, in equilibrium on the fulcrum G. Conceive the lever to be turned intinitesimally about G, fio. 89. taking the position CD, then will Aa, which is the projec- tion of the path A on the action-line of P, be the virtual velocity of P ; and similarly Bh will be the virtual velocity of W. The former will be positive and the latter negative; hence P.Aa - W.Bb = 0. The triangles AaO And AUG at the limit are similar, having the right angles AaO and AOG, aAO = AGO, and the ro- maining angle equal. Similarly, bBB is similar to BGD. 11 162 VIRTUAX VELOCITIES. aoi.i' M = bg = ^^''^^^''^''^bg'^ which, substituted in the preceding expression, gives F.AG = WBG; that is, the weights are inversely jpro^portioned to the arms. If the lever be turned about the end A^ we would find in- a similar manner that {P -f W).AG = W.AB \ in which P + TFis the reaction sustained bj the fulcrum G. 2. Find the conditions of equilibrium of the bent lever. Let ^6^ and GB be the arms of the lever and G the fulcrum. Let it be turned slightly about G ; then will Aa and Bh be the respective virtual velocities of no. w. Pand F; ^ /. - P,Aa + W.IB = 0. From G draw GG perpendicular to PA, then will the tri- angle A CG be similar to AaA\ having the angle AaA' = A CG\ and a A A! = CGA, Similarly, the triangle BDG is similar to BhB'-, Aa _ GG^ •*• Bb- lU)' which, combined with the preceding equation, gives P.GG= W.GD; that is, ths weights are inversely proportional to their horizot^ ted distances from the fulcrum.. 3. Find the conditions of equilibrium of the single pulley. r\ P w Fia. 91. In Fig. 91, let the weight P be moved a distance equal to ah, then will W be moved a distance cd = ah ; hence, we have - P.ab + W.cd =r : ,, P = W, 1101.1 EXAMPLES. 163 4. On the inclined plaiie AG^ 9k weight P is held by a force TF" acting parallel to the plane ; reqnired the relation between /* and W. dez=.ah will be the virtual velocity of TF, and ac that of P ; and we have - P.ac + W,ab = 0. From the similar triangles abc and ABC^ we have Fio. 02. ac ah PR -^ ..P.CB=W.AC', or Fio, AB : BGy P: W :: AC 5. On the inclined plane, if the weight P is held by a force TF, acting horizontally, required the relation between P and TF. Tne movement being made, cd will be the virtual velocity of TF, which at the limit equals ae, and be will be the virtual velo- city of P, and we have — P.be + W,ae = ; and ae : eh ,'.P.CB= W.BA; or, the weight is to the horizontal force as the base of the tri- angle is to its altitude T). In Fig. 27 show that Pdr = Wdy. 7. One end of a beam rests on a horizontal plane, and the other on an inclined plane ; re- quired the horizontal pressure against the inclined plane. This involves the principle c 'f the wedge ; for the block ABC may represent one-half of ^^®- •*• a wedge being forced against the resistance TF. Conceive the pV'*^ to bo moved a distance AA\ and that the beam turns w -^ 164 VIBTUAL VELOCITIES. [lOt.J about the end D^ but is prevented from sliding on the plane j then will the virtual velocity of the horizontal pressure be AA\ and that of the weight will be Eg ; hence, for equilibrium we have W,Eg - P.AA' = 0. (a) We now find the relation between Eg and AA\ Let I = DF, the length of the beam ; a = DE, the distance from D to the centre of gravity of the beam ; a= GAB\ fi=^ ADF. The end at F will describe an arc FF' about i> as a centre. From F' draw F'd parallel to AA\ and from F drop a per- pendicular Fe upon dF', Then, from similar triangles, we have Fe = -J- EC, FF' will be dF, hence perpendicular eFF' = AI)F-- to DF, and Fe perpendicular = ^\ dFe= 90°— a; to and .-. dFF' : = 90°-a + /3; FF' =z Fe^QQS = - Eg sec fi. d The triangle FdF' gives FF' sin a AA' = dF' ' ~ sin (90° _ a + /3)» hence, Eo « sin a cos ff ^ J. J.' Z co8(a-/3)' which, substituted in equation (a) above, gives T) __ -rn- ^ sin a cos fi ~~ I cos (a — /3)* 8. Deduce the formula for the triangle of forces from the principle of Virtual Yelocities. CHAPTEK Vll. MOMENT OF INERTIA. (IIj^b chapter may be omitted until its principles are needed hereafter (bm Ch. X) Although the expression given below, called the Moment of Inertia^ comes directly from the solution of certain mechanical problems, yet its prin- ciples may be discussed without involving the idea of force^ the same as any other mathematical expression. The term probably originated from the idea that inertia was considered a force ^ and in most mechanical problems which give rise to the repression the moment of a force is involved. But the expres- sion is not in the form of a simple moment. If we consider a moment as the product of a quantity by an arm, it is of the form of a Tnomeiit of a moment. Thus, dA being the quantity, i/dA would be a moment, then considering thia as a new quantity, multiplying it by y gives y'^dA^ which would be a moment of the moment. Since we do not consider inertia as a force, and since all these problems may be reduced to the consideration of geometrical magnitudes, it appears that some other term might be more appropriate. It being, how- ever, nniversally used, a change is undesirable unless a new and better one be unifoersaUy adopted.) , DEFINITIONS. 102. The expression, y^^.^, m which dA represents an ele- ment of a body, and y its ordinate from an axis, occurs fre- quently in the analysis of a certain class of problems, and hence it has been found convenient to give it a special name. It is called the moment of inertia, THE MOMENT OF INERTIA OF A BODY is the sum, of the products obtained hy multijplyvng each element of the body hy the square of its distance from an axis. The axis is any straight line in space from which the ordinate is measured. The quantity dA may represent an element of a line (straight or curved), a surface (plane or curved), a volume, weight, or mass ; and hence the above definition aiiswers for all these quantities. 166 MOMENT OF INERTIA. tlOS.J The moment of inertia of a plane surface, when the axis lies in it, is called a rectangular moment; but when the axis is perpendicular to the surface it is called ^ polar moment. 103. Examples. 1. Find the moment of inertia of a rect- angle in reference to one end as an axis. Let h = the breadth, and d = the depth of the rectangle. Take the origin of coor- dinates at 0, We have dA = dydx ; Fio 96. and \hd^. Fio. 96. 2. What is the moment of inertia of a rectangle in reference to an axis through the centre and parallel to one end ? Ans. -^^h(P. 3. What is the moment of inertia of a straight line in reference to an axis through one end and perpendicular to it, the section of the line being considered unity ? - Ans. iZ3. 4. Find the moment of inertia of a circle in reference to an axis through its centre and perpendicular to its surface.^ We represent the polar moment of inertia by ip. Let r = the radius of the circle ; p = the radius vector ; = the variable angle ; then dp = one side of an elementary rectangle ; pdd = the other side ; and dA = pdpdO ; and, according to the definition, we have »27r p^dpdO = i^rr*. J104.i EXAMPLES. 167 5. What is the moment of inertia of a circle in reference to a diameter as an axis? (See Article 105.) AnS. iTTT^. 6. What is the moment of inertia of an ellipse in reference to its major axis ; a being its semi- major axis and ^, its semi-minor? fio. 97. Ans. ^ab^, 7. Find the moment of inertia of a triangle in reference to an axis through its vei-tex and parallel to its base. Let b be the base of the triangle, d its altitude, and x any width parallel to the base at a point whose ordinate is y ; then dA = dxdi/j and we have b 1= f p^faydx=-.^ C fdy = \hd^. 8. What is the moment of inertia of a triangle in reference to an axis passing through its centre and parallel to the base ? Ans, -sjbd^. 9. What is the moment of inertia of an isosceles triangle in reference to its axis of symmetry? A?is. -i^b^d. 10. Find the moment of inertia of a sphere in reference to a diameter as an axis. The equation of the sphere will hQ 7? + y^ -{- £^ =. U^, The moment of inertia of any section perpendicular to the axis of x will be \inf"i hence for the sphere we have FORMULA OF REDUCTION. 104. The moment of inertia of a body^ in reference to any axis, eqiuils the moment of inertia in reference to a jparallel ijLdds passing through the centre of the body jplus the product of the area {or volume or mass) by the sq^mre of the distance between the 0^X68, 168 FORMULA OF REDUCTION. [104.1 This proposition for plane areas was proved in Article 80. To prove it generally, let Fig. 98 re- present the projection of a body upon the plane of the paper, B the projection of an axis passing through the centre of the body, A any axis parallel to it, C the projection of any element; AG -■=■ r^ BO = n, the angle CBD = 6, and V = the volume of the body. ' Then Fio. 98. /i = Jr^^d V will be the moment of inertia of the volume in reference to the axis through the centre ; and I =Jr^dV, the moment in reference to the axis through A, Let AB = D, then Ai: = D -\- r^ cos 6, and t^ = r{^ sin^ $ + {D + r^ cos 6f ; .-. f^dY=fr^^dV + "IDJn cos edV -^ D^fdY. But yVi cos ddV = 0, since it is the statical moment of the body in reference to a plane perpendicular to AD passing through the centre of the body and perpendicular to the plane of the paper, therefore the p^receding equation becomes 1= /,+ VL^\ (119) which is called i\\Q formula of reduction. From this, we have 1^ = 1 -VLf, (120) EXA MPLE S 1. The moment of inertia of a rectangle in reference to one end as an axis being ^d^^ required the moment in reference to a parallel axis through the centre. Equation (120) gives 7, = \U^ - hd {\df = ^U\ 2. Given the moment of inertia of a triangle in reference to an axis through its vertex and parallel to the base, to find the moment relative to a parallel axis through its centre. no5.j EXAMPLES. 169 Example 7 of the preceding Article gives I = \bd^ ; hence equation (1 20) gives 3. Find the moment of the same triangle in reference to the hase as an axis. Equation (119) gives / = ^^hd^ + ihd iid)^ = ^d\ 105, To FIND THE RELATION BETWEEN THE MOMENTS OF IN- KSTIA IN REFERENCE TO DIFFERENT PAIRS OF REOTANGULAB AXES HAVING THE SAME ORIGIN. Let X and y be rectangular axes, Xi and yi, also rectangular, having the same origin ; a = the angle between x and Xi ; 7» = the moment of inertia relatively to the axis X, similarly for /y,/«»and/y,; B = JxydA ; and ^1 =AiyidA, For the transformation of coordinates we have Xi = y Bin a + a cos a ; yi = y cos a — X sin a ; a^2 + y,2 ^ ar» + f. Fia. Wi Also dA = dxdy = dx^dyi. Hence, f^^ = fy^ dA = /jbCOS^ a -\- ly sin^ a —2B cos a sin a ; ly^ z= Tx sin' a -^ ly cos'^ a + 2^ cos a sin a ; Bi = {Ix — ^ ) cos a sin a + j5 (cos' a — sin' a) ; '*. /«, + ^1 ^= /at + ^y — /p> (121) the last value of which is found from the expression fy^dA -h fx^dA=f{jy^ + x^ dA ^fp^dA = 7^; which shows that the 170 MOMENT OF INERTIA. [105.1 ].)olar moment equals the Bum of two rectangular moments, the origin being the same. If the rectangular moments equal one another, we have Ip = 2/a. . Thus, in the circle, Ip = iirr^. (See Ex. 4, Article 103), hence /^ = Jtt/^. The last of equations (121) is an isotrojpic function; since the sum of the moments relatively to a pair of rectangular axes,equals the sum of the moments relatively to any other pair of rectangular axes having the same origin ; or, in other words, the sum of the moments of inertia relatively to a pair of rect- angular axes, is constant. To find the maximum or minimum moments we have, from the preceding equations, da and = ~ (ii — Ip) coa a sin a — B (cos^ a — sin^ a) = ; da = + (^ — ly) cos a sin a -H ^ (cos^ a — sin'* a) = ; .-. JB,= 0. From the first or second of these we have — 2B 2 cos a sin a Iv = tan 2a. It may be shown by the ordinary tests that when 7a:, is a maximum, ly^ will be a minimum, and the reverse ; hence there is always a pair of rectangular axes in reference to one of which the moment of inertia is greater than for any other axis, and for the other it is less. These are called jprincijpal axes. Thus, in the case of a rectangle, if the axes are parallel to the sides and pass through the centre, we find B= I xydA = 0; hence x and y are the axes for maximum and minimma moments ; and if c? > J, yV^<^^ ^^ ^^^ maximum, and -^^^d a minimum moment of inertia for all axes passing through the -origin. In a similar way we find that if the origin be at any [105.] EXAMPLES. 171 other point the axes mnst be parallel to the sides for maximum and minimum moments. The preceding analysis gives the position of the axes for maximum and minimum moments, when the moments are known in reference to any pair of rectangular axes. But if the axes for maximum and minimum moments are known as i^ and ly^ then ^ = 0; and calling these I^ and i^, Eqs. (121) become 4 = /a/ cos^ a -\- lyf sin* a ; = Ig^ sin'* a -\- lyt cos' a ; (122) B^ = (7a/ — Ij/) COS a sin a. In the case of a square when the axes pass tli rough the centre Ix' — ly' \ .-. I^ = I^ (cos' a + sin' a) = I^^; ly^ ■=. lyt^ and 5. = 0; hence the moment of inertia of a square is the same in refer- ence to all axes passing through its centre. The same is true for all regular polygons, and hence for the circle. Examples. 1. To find the moment of inertia of a rect- angle in reference to an axis through its cen- ^ tre and inclined at an angle a to one side, we have /a. = ^^Id^ and ly = j\¥d ,\ /asj = ^\bd (d^ cos' a + P sin' a) ; 7^^ = r^\M (d^ sin' a + P cos' a). no. loo. 2. To find the moment of inertia of an isosceles triangle in reference to an axis through its centre and inclined at an angle a to its axis of symmetry. We have Ze = ^jM^ and ly = ^h^d, in which h is the base and d the altitude ; /. 7^ = ^\hd (^' cos' a + iP sin' a) ly = -^\hd {d^ sin' a + f^ cos' a). The moment of inertia of a regular polygon about an axiB 172 MOMENT OF INERTIA. [106.J through its centre may be found by dividing it into triangle* having their vertices at the centre of the polygon, and for bases the sides of the polygon ; then finding the moments ol the triangles about an axis through their centre and parallel to the given axis and reducing them to the given axis by the for- mula of reduction. If ^ be the radius of the circumscribed circle, r that of the inscribed circle, and A the area of the polygon ; then, for a A B regular polygon, we would find that For the circle R =. r, S s .:I=i7rr\ Fig. 101. as before found. For the square, r = ih,Ji = ih V2, and A = i^; as before found. 106. Examples of the moment of inertia of solids. (The following results are taken from Mosley's Mechanics arid Engineering.) 1. The moment of inertia of a solid cylinder about its axis of symmetry, r being its radius and A its height, is ^Kt^, 2. If the cylinder is hollow, g the thickness of the solid part and B. the mean radius (equal to one-half the sum of the external and internal radii) Fia.102. then /= '^irhcB (W' + \(?\ 3. The moment of inertia of a cylinder in reference to an axis passing through its centre and perpendicular to its axis of y. symmetry is JttA/^ {^ -H -JA^). 4. The moment of inertia of a rectangular paral- ^^^^m lelopipedon about an axis passing through its c^n- rri 1 1 tre and parallel to one of its edges. Let a be the I j II length of the edge parallel to the axis, and b and e t I "nf the lengths of the other edges, then /= -^^ ahc I (5^ + c^) = T^ of the volume inultvplied hy thi Fie. 108. square of the diagonal of the hose. IIOT.J RADIUS OF GYRATION. 173 5. The moment of inertia of an upright triangular prism having an isosceles triangle for its base, in reference I to a vertical axis passing through its centre of ^\J~7 gravity. Let the base of the triangle be a, its altitude J, and the altitude of the prism be A, then \^ / Fio. 104 6. The moment of inertia of a cone in reference to an of symmetry is ^irr^h, {r being the radius of the base and h the altitude.) Pio. 105. Fia. 106. i?m. iu<. . 7. The moment of inertia of a cone in reference to an axis through its centre and perpendicular to its axis of symmetry ia i^irr^h (/^ + iA^). 8. The moment of inertia of a sphere about one of its diam- eters is r^Trl^. 9. The moment of inertia of a segment of a sphere about a diameter parallel to the plane of section. Let H be the radius of the sphere, and b the distance of the plane section from the centre, then / = ^TT (16^ + 15^^ + lOH^b^ - 9^). Y^^7v^. RADIUS OF GYRATION. 107, ^^ inay conceive the mass to be concentrated at such a point that the moment of inertia in reference to any axis will be the same as for the distributed mass in reference to the same axis. The radvas of gyration is the distance from the moment axis to a point in which, if the entire mass be concentrated, the moment of inertia will be the same as for the distributed mass 174 EXAMPLES. [107.] The ^nincijpal radius of gyration is the radius of gyration in reference to a moment axis through the centre of the mass. Let k = the radius of gjj-ation ; h = the principal radius of gyration ; M = the mass of the body ; and D = the distance between parallel axes ; then, according to the definitions and equation (119), we have M^ = Sm7» = Xmr? + ML^ = Mk} + i/Z^; .-. 1^=hi -V iy\ (123) from which it appears that ^ is a minimum, for Z^ = 0, in which case ^ = ^ ; that is, the jprincijpal radius of gyration is the minimwin radius for parallel axes. We have Jience, the square of the principal radius of gyration equals the moment of inertia in reference to a moment axis through the centre of the body divided by the mass. Examples. 1. Find the principal radius of gyration of a circle in refer- ence to a rectangular axis. Example 5 of Article 103 gives, I^ = i^r/**, which is the moment of an area, hence, we use irr^ for Jf, and have .,=§;= ^. 2. For a circle in reference to a polar axis, h^ = ^. 3. For a straight line in reference to a moment axis perpen- dicular to it, k^ = 1^?^- 4. For a sphere, k^ = f r*. 6. For a rectangle whose sides are respectively a and 5, in reference to an axis perpendicular to its plane, 1^=^^ {a^+V), 6. Find the principal radius of gyration of a cono when the moment axis is the axis of symmetry. CHAPTER VIII, XOnON OF A PARTICLE FREE TO MOVE IN ANY DIKECrnON. 108. A free, material particle, acted upon by a system ^f forces which are not in equilibrium among themselves, will describe a path the direction of motion in which immediately after passing any point will depend upon its direction when it arrives at the point and the resultant of the forces acting upon it at the point, but the direction of the acceleration will be in the direction of the resultant of the impressed forces. Let R be the resultant of the forces acting upon a particle whose mass is m at the point whose coordinates are a?, y, 2, and da' the path which would be described by the effect of M only ; then, according to Article 21, we have Let a be the angle between the action-line of the resultant R (or of the arc ds') and the axis of x ; multiplying by cos a, we have R cos a — m-j^ cos a = ; in which R cos a is the Xrcomjponent of the resultant, and according to equation (51) equals X\ or, in other words, it is the projection of the line representing the resultant on the axis of X ; d}s' cos a is the projection of d^s on the axis of oj, and is d^x, Ilcnce, the equation becomes, d'^x _ w-r^ = 0; and similarlyi de a24) 176 MOTION OP A [109.J which are the equations for the motion of a particle along the coordinate axes ; and are also the equations for the motion of a body of finite size when the action -line of the resultant passea through the centre of the mass. They are also the equations of translation of the centre of any free mass when the force? produce both rotation and translation ; in which case m should be changed to M to represent the total mass. See Article 38. VELOCITY Aim LIVING FOEOB, 109, Multiplying the first of equations (124) by dx, the second by dy, and the third by dz, adding and reducing, give Xdx + Ydy + Zdz — imd( '—^ j = imd -^ ; and integrating gives f{Xdx + Ydy + Zdz) = im^ = imv^ + O, The first member is the work done by the impressed forces ; for if ^ be the resultant, and s the path, then, according to Article 25, equation (26), the work will be /I^ds, and by pro- jecting this on the coordinate axes and taking their sum, we have the above expression. The second member is the stored energy plus a constant. Let X, y, Z be known functions of x, y, 2, and that the terms are integrable. (It may be shown that they are always integrable when the forces act towards or from fixed centres.) Performing the integration between the limits a?^, y^, s^, and «Ji, yi, ^, we have </> (^oi Voy ^0) - </> (^j yiy ^1) = 4^ (^0 - -2^1) ; (125) hence, the work done by the impressed forces upon a body in passing from one point to another equals the difference of the living forces at those points. It also appears that the velocity at two points will be independent of the path described ; also, that, when the body arrives at the initial point, it will have the (Same velocity and the same energy that it previously had at that point. iio».| FREE PABTIOLK. 177 Examples. \,Ifa body iajprojected into space^ arid acted upon only by gravity and the im/pulse / required the cv/rve described by ths jyrqjectile. Take the coordinate plane xy in the plane of the forces, x horizontal and y vertical, the origin being at the point from which the body is projected. Let W = the weight of the body ; V = tiie velocity of projection ; and j. a = BAX = the angle of elevation at which the projection is made. We have, X = 0; r=-7^; Z=0; and equations (124^ become Via. 109. 5 = 0; Integrating, observing that v cos a will be the initial velocity along the axis of a?, and v sin a that along y, we have, dx -— = V cos a : dt ' dy -4 = v%ina dt gt\ and integrating again, observing that the initial spaces are zero we have, X =^ vt cos a ; ] y = vt m\ a — igft\ ) Eliminating t from these equations, gives which is the equation of the common parabola, whose axis is parallel to the axis of y. 12 178 PROJECTILES. lM.y Let h be the height through which a body must fall to acquire a velocity v, then tj^ = ^gh, and the equation (b) becomes, Tojlnd tJie range AE^ make y = in equation (J), and we find a? = 0, and X = AE = 4A cos a sin a = 2h sin 2a ; {d) which is a maximum for a = 45°. The range will be the same for two angles of elevation, one of which is the complement of the other. The greatest height^ will correspond to a? = A sin 2a, which, substituted in (J), gives, Asin^a. («) The velocity at the end of the time t ifi. or by eliminating t by means of the first of equations (a\ we have, The direction of motion at any point is found by differentiating equation (c), and making tan ^ — 3^ = tan a — —, 5- . (A) dx 2Acos a ^ ' At the highest point — 0, .\ x = h sin 2a, as before found. For OJ = 2A sin 2a, we have, tan 6 =. — tan a, or the angle at the end of the range is the supplement of the angle of projection. 2. A body is projected at an angle of elevation of 45°, and has a range of 1,000 feet ; required the velocity of projection,, the time of flight, and the parameter of the parabola. lOa.J EXAMPLES. 179 3. What must be the angle of elevation in order that the horizontal range may equal the greatest altitude ? What, that it may equal n times the greatest altitude? ' 4. Find the velocity and the angle of elevation of a projectile, 8o that it may pass through the points whose coordinates are flJi = 400 feet, y^ = 60 feet, x^ = 600 feet, and ya = 40 feet. 5. If the velocity is 600 feet per second, and the angle of elevation 45 degrees ; required the range, the greatest elevation, ^e velocity at the highest point, the direction of motion 6,000 feet from the point of projection, and the velocity at that point 6. If a plane, whose angle of elevation is i, passes through the origin, find the coordinates of the point C, Fig. 109, where the projectile passes it. 7. In the preceding problem, if * is an angle of depression, find the coordinates. 8. Find the equation of the path when the body is projected horizontally. 9. If a body is projected in a due south- erly direction at an angle of elevation a, and is subjected to a constant, uniform, horizontal pressure in a due easterly direc- tion ; required the equations of the path, neglecting the resist- ance of the air. We have X = ; Y= ^ mg ; Z = F{^ constant). The projection of the path on the plane osy will be a para- bola, on xz also a parabola. 10. If a body is projected into the air, and the resistance of the air varies as the square of the velocity ; required the equa- tion of the curve. (The final integrals for this problem cannot be fonnd. Approximate solu- tions have been made for the purpose of determining certain laws in regard to goimery. It is desirable for the student to establish the equations and make the firnt steps in the reduction. ) ( rhe remainder of this chapter may be omitted without detriment to what follows it. It, however, contains an interesting topic in Mechanics, and iB of ▼ital importance in Mathematiocd Astronomy and Physics. ) 180 CENTRAL [110,111.1 CENTRAL FORCES. 110, Central forces are such as act directly towards or from a point called a centre. Those which act towards the centre are called attractive^ and are considered negative, while those which act from the centre are repulsive, and are considered positive. The centre may be fixed or movable. The line from the centre to the particle is called a radius vector. The path of a body under the action of central forces is called an orbit. The forces considered in Astronomy and many of those in Physics, are central forces. GENERAL EQUATIONS. Ill, Consider the force as attractive, and let it be represented hy-K Take the coordinate plane xy in the plane of the orbit, the origin being at the centre of the force, and OP = /•, the radius vector, then X = — i^cos a ^p/ ML Fia. 111. r=-i5^cos)S=-i^-^; and the first two of equations (124) become dp /•' m fK (126) To change these to polar coordinates, first modify them By multiplying the first by y and the second by x, and subtracting, and we have d^x dJ^y 0; fll2.] FORCES. 181 and multiplying the first by x and the second by y, and adding we have Let Q — POM— the variable angle, then aj = /• cos ^, y = /• sin ^, and differentiating each twice, we find dj'x = (di'v - rde^) cos d — i^d/rdd + rd'^B) sin ^; dJ^y = (d^T -^dO^) sin 6 + i^drde + /•<^2^) cos B ; which substituted in the preceding equations, give dj'r (dd\^ F „^^ df drdd dJ'd ^ df ' 6?^ which may be put under the form Idi^dd m-^H- <»> Equation (127) shows that the acceleration along r is the force on a unit of mass; and (128) shows that there is no acceleration perpendicular to the radius vector. 112. Principle of equal areas, — Integrate equation (128), and we have ^f =^; (129) , and integrating a second time, we have fi»dQ = Ct ; (130) the constant of integration being zero, since the initial values of t and Q are both zero. But from Calculus y7^<^^ is twice the sectoral area POX\ hence the sectoral area swept over by the radius vector increases directly as the time ; and equal a/reM urill he passed over in equal times. 182 CENTRAL [113.1 Making ^ = 1, we find that G will be twice the sectoral area passed over in a unit of time. The converse is also true, that if the areas are proportional to the times the force will be central. For, multiplying the first of (126) by y and the Becond by x and taking their difference, we have or, Xy- Ja; = ; which is the equation of a straight line, and is the equation of the action-line of the resultant, and since it has no absolute term it passes through the origin. 113. To find the equation of the orbit^ diminate dt from equations (127) and (128). For the sake of simplifying the final equation, make /* = -, and (129) becomes Differentiating and reducing, gives du ^dudt <^=-^ = -^'^ ^'» dr _ du dt" ^dd' the first member of which is the velocity in the direction of the radius. Differentiating again, gives d^- ^"^ d^' Substituting these in equation (127) and at the same time making m equal to unity, since the unit is arbitrary, we have 1114.1 FORCES. 183 or, cPu F ^ which is the differential equation of the orbit. When the law of the force is known, the value of F may be substituted, and the equation integrated, and the orbit be definitely determined. Multiplying by du and integrating the first two terms, we have 114. Given the equation of the orbit, to find the law of the force. From equation (132), we have ^= ^MS + 4 <i^) Another expression is deduced as follows : let p = the perpendicular from the centre on the tangent, then from Calculus we have ^ ^ di ^ dr' + r'de^ " ^^TZ' (135) Differentiating, gives d^u -w""" i>^jp = -nr:^ 73^^; tS4 CENTRAL [115.) and dividing by ^*, substituting, du — — uHr^ and reducing^ we have 1 d/p ^(dJ^u \ which, combined with equation (134), gives which is a more simple formula for determining the law of the central force. 115. To determme the velocity at any jpoint of the orbit. We have __d8 _ds dd __ ds dd ^^di~"di dd ~dedi ds = Ou^ -^ (from equation (131) ) = -. (from Bif, Oal), (137) Hence, the velocity va/ries inversely as the perjpendicula/t from the centre v{pon the tangent to the orbit. Another expression is found by substituting the value of p^ equation (135), in (137). Hence, Idu" '0= CSJ-^ ^u\ (138) Still another expression may be found by substituting equa- tion (138) in (133) ; hence v^^ C, + 2fF^ (139) = (7i - ^fFdr, Since -?^ is a function of r, the integral of this equation gives V in terms of r, or the velocity depends directly upon the n 16, 117 1 FOECES. 185 distance of the body from the centre. Hence, the velocity at any two points in the orbit is independent of the path between them, the law of the force remaining the some. 116, To determine the time of describing any jportion of the orbit. To find it in terms of r, eliminate dQ between equations (129) and (133), reduce and find -P "^ (1*0) which integrated gives the time. To find it in terms of the angle, we have from (129) Ur^dS, C from which r must be eliminated by means of the equation of the orbit, and the integration performed in reference to 6. 117, To find the com/ponenta of the force along the tanqewt and normal. Let T = the tangential component ; N = the normal component ; and resolving them parallel to x and y, we have dr da d%* ^5^«= ^^^ds^^di illliminating iT gives d'^x ^ dhj ^ ^^ m^dx^m-^dy^TdBi or dx . ^dy , €Ps 186 CENTRAL FOBCBS. |117.l Eliminating 2* gives ,^ dx dJhj dy (Px da d^ ds d^ __ md^ idxdj^y — dyd}x\ ~ d^ds \ d^ ■/ = ^^ ; (141) hence the component of the force in the direction of the nor^ mal is dependent entirely upon the velocity and radius of cur- vature. This is called the centrifugal force. It is the measure of the force which deflects the body from the tangent. The force directed towards the centre is called centripetal, . If (u = the angular velocity described by the radius of cnp- vaturo, then, v = />©, and equation (141) becomes N = m^p. (142) Examples. 1. If a body on a smooth, horizontal plane is fastened to a point in the plane by means of a string, what will be the num- ber of revolutions per minute, that the tension of the string may be twice the weight of the body. 2. A body whose weight is ten pounds, revolves in a hori- zontal circle whose radius is five feet, with a velocity of forty feet per second ; required the tension of the string which holds it (Use equation (141).) 3. Eequired the velocity and periodic time of a body re- volving in a circle at a distance of n radii from the earth's centre. The weight of the body on the surface being mg, at the dis- tance of n radii it will weigh mg ( — 1 = — |^, and this is a {117.] EXAMPLES. 187 measure of the force at that distance. (Use equation (141) or (139).) ^--=(f^' = ^'^(T)* (This is snbstantially the problem which Sir Isaac Newton used to piOT* the law of Universal Gravitation. See Whewell's Inductive Sciences.) 4. A jpariide is projected from a given point in a given direction with a given velocity y and moves under the action of a force which varies inversely as the square of the distance from the centre / required tJie orbit. Let /A = the force at a unit's distance, then F= fiu\ and equation (132) becomes d^u fi ^ cr, the first integral of which becomes by reduction de = in which ^ is an arbitrary constant, and the negative value of the radical is used. Integrating again, making 0q the arbitrary constant, we have which by reduction gives 188 ORBITS OF [1170 which is the general polar equation of a conic sectioD, the origin being at the focus. As this is the law of Universal Gravitation, it follows that the orbits of the planets and comets are conic sections having the centre of the sun for the focus. In equation {a), 6q is the angle between the major axis and a line drawn through the centre of the force, and is the eccentricity = e ; hence the equation may be written u = -^ (l + ^ cos (^ ~ ^o)). W The magnitude and position of the orbit will be determined from the constants which enter the equation, and these are determined by knowing the position, velocity, and direction of motion at some point in the orbit. Draw a figure to represent the orbit, and make a tangent to the curve at a point which we will consider the initial point. Let fi be the angle between the path and the radius vector at the initial point, Vq the initial radius vector, and Vq the initial velocity; then at the initial point «_1, ^ = 0, cot5 = ^J = du (") id from equation (J) 0^ e C0& 6o =. 1, [d) du /^ . /I liich, combined with equation (c), gives = — d sm ^0, (fi) From equation (137) 6^= Forosin^S; (/) tI17.] THE PLANETS. 18» which, substituted in equations (d) and («), and the latter divided by the former, gives /) y^o\ sin «os 13 tan Un = tjts . o A» /A — FoVo sin* y8 Squaring equations (cT) and («), adding and reducing by eqna* tion (/), give Hence, when 2/i, Fo' > — , « > 1, and the orbit is a hyperbola, 2lJL Fq' = — , 6 = 1, and the orbit is a parabola, 2it Fo' < — , 6 < 1, and the orbit is an ellipse. or (see example 23, page 34) the orbit will be a hyperbola, a parabola, or an ellipse, according as the velocity of projection is greater than, equal to, or less than the velocity from infinity. As the result of a large number of observations upon the planets, especially upon Mars, Kepler deduced the following laws : 1. The planets describe ellipses of which the Sun occupies a focus. 2. The radius vector of each planet passes over equal areas in equal times. 3. The squares of the periodic times of any two planets are as the cubes of the major axes of their orbits. The first of these is proved by the preceding problem, since the orbits are reentrant curves. The second is proved by equation (130). The third we will now prove. 5. Required the relation between tlie time of a comjplete circuit of a pa/rticle in cm ellipse^ and the major axis of the 07'hit, Let the initial point be at the extremity of the major axis near the pole, then 6^ in equation {d) will be zero, and we have t7' = /*ro(l + e); 190 CENTRAL FORCES. fll7 ] but from the ellipse, r^ = a — ae = a (1 — e); Equation (180) /. (7 = V^ui(l-^. gives ^ 2 area of ellipse ^ ~ G 27ra%/(l -<5») V fiail- V ;t -^ 6. The orbit being an ellipse, required the law of the force The polar equation of the ellipse, the pole being at the focus, IB __1_ 1 + g cos (^ - ^o) ^ "■ r "" a (1 - ^) ' which, differentiated twice, gives dhi __ e cos (6 — 0^ e?^ "" a{l-e^ ' and these, in equation (132), give, ^ "■ « (1 - e^) 7^ ' hence the force varies inversely as the squa/re of the dista/nce, 7. Find the law of force by which the particle may describe a circle, the centre of the force being in the circumference of the circle. (Tait and Steele, Dynamics of a Particle,) Am, F^ 4- IT ^ 8. If the force varies directly as the distance, and is attrac- tive, determine the orbit. (This is the law of molecular action, and analysis shows that the orbit is an ellipse. The problem is of great importance in Physice especially in Optioi and Acoustics.) CHAPTER IX OONSTEAINED MOTION OP A PABTIGLB. 118. If a body is compelled to move along a given fixed curve or surface, it is said to be constrained. The given curve or surface will be subjected to a certain pressure which will be normal to it. If instead of the curve or surface, a force be substituted for the pressure which will be continually normal to the surface, and whose intensity will be exactly equal and opposite to the pressure on the curve, the particle will describe the same path as that of the curve, and the problem may be treated as if the particle were free to move under the action of this system of forces. Let N = the normal pressure on the surface, and Z —f{x, y, z) = 0, be the equation of the surface ; 6x, Op, d» the angles between -A^and the respective oodr- dinate axes. Then X -{■ Hfco&Ox—m-j^ = 0; T + iTcos dy— m df Z -h iTcos 0» — m-j^ dr 0; (US) in which the third terms are the measures of the resultants of the axial components of the applied forces. We will confine the further discussion to forces in a plane. Take ocy in the plane of the forces, then we have d'^x ^ ^^dy de do (144) •Stt 192 CONSTRAINED MOTION OP flW.l Eliminating N, we find m \ — ^^ + -^ \ = Xd^+ Ydy, or m it)'-<i)'\''^'^*'r^: *and integrating, making v^ the initial velocity along the path, and V the velocity at any other point, we have 4m («r» - v,") =f{Xdx + Ydy) ; (145; hence, the living force gained or lost in passing from on6 point to another is equal to the worJc done hy the vm/pressed forces. If the forces JT and Y are functions of the coordinates X and y, and the terms within the parenthesis are integrable, the result may be expressed in terms of constants and functions of the coordinates of the initial and terminal points, and may be written ^m{^ — v^) = c^ (a?o,yo) — ^ <^ {^^V) ; hence, for such a system, the velocity will he independent of the fath described^ and will he dependent only ujpon the coordinates of the points ; also, the velocity will he independent of the nor- mal pressure, 119. To find the normal pressure multiply the first of equations (144) by dy, the second by d^, subtract, observing that da? ^- d^ = d^, and we find ) dx d^y dyd^x}_ ^dx i^'^y , 7\r. "^ydsTf'dslfl-^ds ^ds^"-"' or ds^ I dxd^y — dyd^x \ _ y^ — X^ ^ N- '^ df\ di X ds ds ' ,,B=X^^-Y^^m,t. (146)' ds ds p ^ ^ m which p is the radius of curvature at the point. The first and second terms of the second member are the normal com- ponents of the impressed forces. The total normal pressiire liaO, 121.J A PARTICLE. 198 unU^ therefore^ he that due to the imjp^ eased forces plus that dice to the force necessanry to deflect the body from the tangent. The last term is called the centrifugal force, as stated in Article 117. If the body moves on the convex side of the curve, the last term should be subtracted from the others ; hence it might be written ±m—; in which + belongs to movement on the concave arc, and — on the convex. 120. To flnd the time of movement^ from equation (145), we have Vmds -J (1*1) 121. To flnd where the particle will lea/ve the constrain- ing curve. At that point N=0, which gives P -^ = ^1-^^' <i«) which, combined with the equation of the curve, makes known the point. If a body is subjected only to the force of gravity, we have X = in all the preceding equations. Examples. 1. A body slides down a smooth inclined plane under the force of gravity / required the formulas for the miction. Take the origin at the upper end and let the equation of the plane be 13 194 EXAMPLES. IH1.» y being positive downward. Then we have X = 0, Y= 7ng, dy = adx^ v^ — 0, and equation (145) becomes i;'* = 'ifgadx = 2gax = 2ffy ; (a) hence, the velocity is the same as if it fell vertically through the same height. To find the time, equation (147) gives that is, if the altitude of the jplane (y) is constant the time varies directly as the lengthy s. "We may also find s = t Vigy = ig^ sin a, (e) 2. Prove that the times of descent down all chords of a ver- tical circle which pass through either extremity of a vertical diameter are the same. 3. Find the straight line from a given point to a given in- clined plane, down which a body will descend in the least time, 4. The time of descent down an inclined plane is twice that down its height ; required the inclination of the plane to the horizon. 6. At the instant a body begins to descend an inclined plane^ another body is projected up it with a velocity equal to tha velocity which the first body will have when it reaches the foot of the plane ; required the point where they will meet. 6. Two bodies slide down two inclined lines from two given points in the same vertical line to any point in a curve in the Bame time, the lines all being in one vertical plane ; required the equation of the curve. 7. A given weight, P, draws another weight, W, up an in- clined plane, by means of a cord parallel to the plane ; through what distance must jP act so that the weight, W, will move * feet after P is separated from it. ri«.i SIMPLE PENDULUM. 195 8. Required a cui've such that if it revolve with a uniform a/ngular velocity about a vertical diameter, T and a smooth ring of infinitesimal diameter \ J y he jplaced ujponit at any jpoint, it will not \ ^^^^^JB slide on the curve. \ / Let 0) be the angular velocity, then we * ^r^ have Fio. lis. Y = — m^, X = mo?x, «> = 0, Vp = 0» and equation (145) becomes ft) V - 2^y + 67 = 0, which is the equation of the common parabola. If the origin be taken at B, G will be zero. (Another Solution. — Let NR be a normal to the curve, MR = the cen- trifugal force, NM = the force of gravity ; but the latter is constant, hence NM, the 'subnormal, is constant, which is a property of the common parabola.) 9. Find the normal pressure against the curve in the pro- ceding problem. 10. The Pendulum. — Mnd the time of oscillation of the simple pendulum,. This is equivalent to finding the time of descent of a particle down a smooth v^ arc of a vertical circle. _ Take the origin of coordinates at A, the lowest point. Let the particle start at Z>, at a height AC =■ A ; when it has arrived at P, it will have fallen through a height OB = h — y, and, according to equation (a) on the preceding page, will have a velocity da X5 tD A Fio. 118. The equation of the arc is flj' = 2ry — j^; (-») henoe d^ = _ (r-yf 2vy — y* <//. 196 EXAMPLES OF fltLJ But da = — ^^y V^y — ^ Considering this as negative, since for the descent the arc is a decreasing function of the time, we have from (a) y) ^ry - f) This may be put in a form for integration by Elliptic Func- tions ; but by developing it into a series, each term may be easily integrated. In this way we find by means of which the time may be approximated to, with any degree of accuracy. When the arc is very small, all the terms containing ^ will be small, and by neglecting them, we have for a complete oscillation (letting I be the length of the pendulum), ^ 9 that i's>^for very small arcs the oscillations may he regarded as isochronal^ or performed in the same time. For the same place the times of vibration a/re di/rectly a^ ih6 9qua/re roots of the lengths of the pendulums. For any pendulum the times of vibration vary inversely as the square roots of the force of gravity at different places. If ^ is constant ^Qc g, 11. What is the length of a pendulum which will vibrate three times in a second ? 12. Prove that the lengths of pendulums vibrating during the [131.] CONSTRAINED MOTION 197 same time at the same place, are inversely as the squares of the number of vibrations. 13. Find the time of descent of a particle down the arc of a cycloid. The differential equation of the curve referred to one end as an origin, x being horizontal and y vertical (r being the radius of the generating circle), is dx = -- — - ^ dy, ^2ry^f ^ 'vAj Ans. " 9 The time will be the same from whatever point of the curve the motion begins, and hence, it is called tautochronaZ, 14. In the simple pendulum, find the point where the tension of the string equals the weight of the particle. 15. A particle is placed in a smooth tube which revolves horizontally about an axis through one end of it / required the equation of the curve traced by the jparticle. The only force to impel the particle along the tube is the centrifugal force due to rotation. Letting r = the radius vector of the curve ; 7*0 = the initial radius vector ; (o = the uniform angular velocity ; we have which, integrated, gives hence, the relation between the radius vector and the arc de- scribed by the extremity of the initial radius vector, is the same fts between the coordinates of a catenary. (See equatic^n {k)y p. 134.) 16. To find a curve joining two points down whioh a par- ticle will slide by the force of gravity in the shortest time. The curve is a cycloid. This problem is celebrated in the 198 CENTRIFUGAL FORCE [123, 123. f history of Dynamics. The sohition properly belongs to the Calculus of Variations, although solutions may be obtained by more elementary mathematics. Such curves are called Bra- ehistochrones. PBOBLEMS PERTAINING TO THE EARTH. 122. To find the value of g. We have, from example 10 of the preceding Article, Making 7' = 1 second and I = 39.1390 inches, the length of the pendulum vibrating seconds at the Tower of London, we have for that place, g = 32.1908 feet. The relation between the force of gravity at different places on the surface of the earth is given in Article 19. The determination of I depends upon the compound pendu- lum. 123. To find the centrifugal force at the equator. We have, from equation (142), for a unit of mass, f=a>'R~B', {a) in which ^, the equatorial radius, is 20,923,161 feet; T, the time of the revolution of the earth on its axis, is 86,164 seconds, and TT = 3.1415926. These values give /= 0.11126 feet. The foKce of gravity at the equator has been found to be 32.09022 feet (Article 19) ; hence, if it were not diminished by the centrifugal force, it would be G = 32.09022 + 0.11126 = 32.20148 feet, and / 0.11126 ^J^^ ^32.20148 289 ■' ' 1124.1 ON THE EAETH. 199 hence the centrifugal force at the equator is y|^ of the on- dinuDished force of gravity. Example. In what time must the earth revolve that the centrifngaJ force at the equator may equal the force of gravity ? Ans. ^ of itsjpresent tiTne, 124, To find the effect of the centrifugal force at differerU latitudes on the earth. Let Z = POQ = the latitude of the point P ; E =z OQ = OP = the radius of the earth; then will the radius of the parallel of lati- tude PP' be p /::^— ^^>V B^^R cos Z. \^;;;;i^^^ The centrifugal force will be in the \^ J plane of motion and may be represented pio. 114. by the line P/*, or Pr =fx = (o^Pi = u^R cos Z ; therefore, the centrifugal force varies directly as the cosine of the latitude. But the force of gravity is in the direction PO, Resolving Pr parallel and perpendicular to PO^ we have Pjp = co^R cos^ Z = j\-g O cos'* Z ; Pg = (i?R cos Z sin Z = -^ G sin 2Z ; the former of which diminishes directly the force of gravity, and the latter tends to move the matter in the parallel of lati- tude PP\ toward the equator. Such a movement has taken place, and as a result the earth is an oblate spheroid. In the present form of the earth the action-line of the force of gravity is normal to the surface (or it would be if the earth were homo- geneous), and hence, does not pass through the centre O, except on the equator and at the poles. The preceding formulas would be true for a rigid homogeneous sphere, but are only approximations in the case of the earth. OHAPTEK X. VOBGES IN A PLAIO: PEODUOING EOTATTON. .»•'' 125, AnGTTLAB motion of a particle about a fixed AXISi Let the body (7, on the horizontal arm AB^ revolve about the vertical axis ED, Consider the body reduced to the centre of the mass, and the force Fx applied at the centre and acting continually tangent to the path described by the particle. This may be done as shown in Fig. 121, In this case the force will be measured in the same way as if the path were rectilinear, for the force is applied along the path. Hence, according to equation (21), / F, m df in which s is the arc of the circle. If ^ be the angle swept over by the radius, and Vi the radius ; then « = n ^, ds = Ti dd, d^s=nd^e; which, substituted in the equation above, gives d^6 df A. (160) If a force, F, be applied to the arm AJSy at a distance a from [128, 137.] ANaULAB AGGELEBATIOIT. 201 tlie axis ED^ producing the same movement of the mass C^ we have, from the equality of moments, Article 65, and the value of F^ deduced from this equation anbatitnted in the preceding one, gives that is, the angular accderation produced by a force, F^on a j>article, m, equals the moment of the force divided hy the rruh ment of inertia of the mass, (For moments of inertia, see Chapter YIL) We observe that, when the force is applied directly to the particle, it produces no strain upon the axis, but that it does when applied to other points of the arm. In both cases there will be a strain of due to centrifugal force, o) being the angular velocity, 126, AlIGULAR MOTION OF A ,-'"'""" ""^-^^ FINITE MASS. Let a body, AB, turn about a vertical axis at A, under the action of constant forces, F, acting horizontally ; 772^, ^2, etc., masses of the ele- ments of the body at the re- 8])ective distances r^, 7*2, etc., Pio. ii«. from the axis A ; and consid- ering equation (151) as typical, we have dJ^d ^Fa XFa ^^ ~~ m^r^-{- m^ri + m^r^ + etc., "~ Sma^ _ moment of the forces /'1K0\ fnorrbent of inertia ' ^ ^ 127. Energy of a rotating mass. Multiply both membem FORMULAS FOR [128, 129.] of the preceding equation by dd, integrate and reduce, and w^ find i^m/^gy =fF.Tdd', (153) in which rdd is an element of the space passed over by F, and F.rdd is an element of work done by i^; hence the second member represents the total work done by i^upon the body, all of which is stored in it. Therefore, the energy of a body rotat- ing about an axis equals the Tnoment of inertia of the masH multijplied hy one-half the square of the angular velocity. If the body has a motion of translation and of rotation at the same time, the total energy will be the sum due to both motions ; for it is evident that while a body is rotating a force may be applied to move it forward in space. Article 38, and that the work done by this force will be independent of the rotation. If -v = the velocity of translation of the axis about which the body rotates ; Q) = the angular velocity ; and I^ — the moment of inertia of the mass; then the total work stored in the body will be iMv^ + ilr^co^ (154) 128. -Ajt impulse. Multiply both members of equation (152) by dt, integrate, and we find d^ _ _ XafFdt But according to Article 27, fFdt is the measure of an im» pulse and is represented by Q, hence Qa moment of the impulse moment of inertia (155) 129. The time required to pass over n circumferences wiU be, in the case of an impulse, 1130, 131.] ANGULAR ROTATION. 208 _^ 2n7r __ 2n7rl 0) ~' Qa In the case of accelerated forces the time will be found b} integrating equation (153). 130. Simultaneous movement of rotation and of transla- tion. If the body be unconstrained, the motion of transla- tion of the body will be that of the centre of the mass. If the axis of rotation is rigid, it may be located anywhere in the body, or even without the body by considering it as rigidly connected with the body, in which case the motion of translation will be that of some point of the axis. In either case the motion of translation may be considered as resulting from a force acting directly upon the axis of rotation, and the rotation, by a force acting at some other point. The two motions may then be considered as existing independently of each other. 13L Formulas for the movement of a body involving both translation and rotation. The general equations for this case are (164) and (165) in the next chapter. In this Article let the rotation be about an axis parallel tos, and the centre of the mass move in the plane xy. Resolve the forces into couples and forces applied at the origin of coordinates, as in Article 83 ; then will the third of equations (86) be the impressed forces which produce rotation. Let R be the resultant of the forces at the origin at any instant, (see Article 84), and 8 be measured along the path described by the point of intersection of the axis of rotation with the plane xy. Then will the principles of Article 21, give i?-J^g = 0; J(Xy-r.)-:j(m,g-^g) = 0; (156) The expression X (Xy — Yx) is the sum of the moments of the impressed forces = ^Fa (Article 60). Transforming the second term of the last equation into polar coordinates having the same origin, we have ^/ d?x cPy\ ^ ^(Pd 204 REDUCED MASS. ri8J.> heDce^ the equations become m df dW df (167) 132. Reduced mass. A given mass may be concentrated at such a point, or in a thin annuhis, that the force or impulse will have the same effect upon it as if it were distributed. To accomplish this it is only necessary that S'mr^ in the second of (15 Y) should have an equivalent value. Let Mhe the mass of the body, k the distance from the axis to the required point, then Smr" = M^, in which k is the radius of gyration, as defined in Article 107. But any other point may be assumed, and a mass determined Buch that the effect shall be the same. Let k be the distance to that point (or radius of the annulus), and J/i the required mass then we have M, Ml, a68) which is called the reduced mass. Examples. 1. A slender har AB, falls through a height A, retaining its horizontal position until one end strikes a fixed obstacle C ; required the angular velocity of the piece and the linear velo- city of the centre iinmediately after the impulse. Let M be the mass of the bar, I its length, V the velocity of the centre at the instant of impact, and v^ the velocity of the centre immediately after impact. Consider the bodies as perfectly non-elastic; then will the effect of the impact be simply to sud- denly arrest the end A. The bar will rotate about a horizontal axis through the centre, as shown by Article 38 ; and, as shown by Articles 27 and 38, the ^ ^ FiQ. in. fiaS.] EZAMPLESb 205 impulse will he Q = H (v — v^); that is, it is the change of velocity at the centre mvUipUed hy the mass. The impact will entirely arrest the motion of the end, A^ at the instant of the impact, and hence at that instant the angular velocity of A in reference to G will be the same as G in reference to A. Equation (155) gives _ moment of impulse ~ m^oment of inertia _ M {v- V,) jl I ' But at the instant of the impact Vi = il(o, solving these give 0) = f I", Vi = iv. We now readily find Q = IMv. To find the velocity of any point in a vertical direction at the instant of the impact, we observe that it may be considered as composed of two parts ; a linear velocity Vi downward, and a right-handed rotation. The actual velocity at A due to rota- tion will be ^l<o = ft;, which will be upward, and the linear velocity downward will be t>i = fy, hence the result will be no velocity. Similarly, the velocity at B wUl be Jt; + i^ = Jv. Also, for any point dia- tant X from G^ we have at the left of G iv — (OX = Wl — 2 j-J; and to the right of G we have 1.(1 + 29. When the bar comes into a vertical position, we easily find 206 ANGULAR MOVEMENT. [132.] that A has passed below a horizontal through C, Every point, therefore, has a progressive velocity, except the point A, at the instant of impact. After the impact the centre will move in the same vertical and with an accelerated velocity, while the angular velocity will remain constant. 2. Suppose that impact takes place at one-quarter the length from A, required the angular velocity. 3. At what point must the impulse be made so that the velocity of the extremity £ will be doubled at the instant of impact ? 4. An inextensible string is wound around a cylinder, and has its free end attached to a fixed point. The cylinder falls through a cer- tain height (not exceeding the length of the free part of the string), and at the instant of the im- pact the cord is vertical and tangent to the cylinder ; all the forces being in a plane ; re- quired the angular velocity produced by the impulse, and the momentum. A71S. f - ; Q = iMv, Fio. iia j\Q-M v: 6. In the preceding problem, let the body be a homogeneous sphere, the string being wound around the arc of a great circle. 6. A homogeneous jprismatto. har AB^ in a horizontal posi- tion constrained to revolve about a vertical fixed axis A, receives a direct imjpulse from a sphere whose momentum is Mv / re- quired tJie angular velocity of the har. The momentum imparted to the bar will depend upon the elasticities of the two bodies. Consider them perfectly clastic. The effo^.t of the impulse will be the same as if the mass of the bar were concentrated at the extremity of the radius of gyration ; hence an equivalent mass at the point G may be determined. Fig. 119. ri82.] EXAMPLES. 207 Let Ml = the mass of AB ; M2 = the reduced mass ; Vi = the velocity of the reduced mass after impact ; a = Aa Then, by equation (158), the mass of the bar reduced to the point C\ will be a* by equation (40) the velocity of M2 after impact will bo __ 2M _ 2Md!' ^^ M + M^""- Ma^ + MJ^'^' W hence, the momentum imparted will be ,^ 2MMJ^ and the moment will be According to equation (155), we have _ morrient of the impulse moTnent of inertia ^MMJc'a V _M<£+M£ _ 23fa This result is the same as that found bj dividing equation (a) by a, as it should be. 7. JSujfpose, in the preceding problem^ that there is no fixed axis, but that the body is free to translate y find where the im- pact must be made t/iat t/ie initial velocity at the etid A shall he zero. 808 ANGULAR MOVEMENT. flSa) Let Mv be the impulse iniparted to the body ; Mk} = the principal moment of inertia ; h = the distance from the centre of the bar to th« required point ; then __ moment of the ifnpulae ~" mom^ent of mertia _ Mvh _ vh and the movement at A in the circular arc will be , , vlh and the initial linear movement will be vlh which, by the conditions of the problem, will be zero; hence vlh or, 2^^ = ^ The distance from A will be A + 4^ = ^±ii^. If the bar be of infinitesimal section .-. h + il = iL The result is independent of the magnitude of tlie impulse. From (l) we have h{il} = h'; hence, h and il are convertible, and we infer that if the im- pulse be applied at A the point of no initial motion wdl be aft tlik..j AXIS OF SPONTANEOUS ROTATION. 209 the point given by equation (5), where the impact was previ- ously applied. 8. In the "preceding jproblem find where the im/puhe raust bs applied so that the point of no initial velocity shaU be at a distance h' from the centre. The initial linear velocity due to the rotary movement found from ifl) of the preceding example, will be and the initial movement of the required point being zervi, we have A A = -^'. (159) If the point of impact be at b, the point a, where the initial movement is zero, will be on the other a s t7 side of the centre of the body. Let B * ^ ' i ' be the centre, then ^'*- ^' h = bB, N = aB, and from (159), we have hh' = bB,aB = h^', (160) and as ^ is a constant, the points a and b are convertible. AXIS OF SPONTANEOUS ROTATION. 133. In the preceding problem the initial motion would have been precisely the same if there had been a fixed axis through a perpendicular to the plane of motion, and hence the initial motion may be considered as a rotation about that axis. If a fixed axis were there it evidently would not receive any shock from the impulse. The axis about which a quiescent body tends to turn at the instant that it receives an imptdso is called the aaois of span- ianeous rotation, 14 JMO mSTANTANBOUS AXIS. flSi^lSit CENTRE OF PERCUSSION. 134. When there is a fixed axis and the body is so stnick that there is no impulse on the axis, any point in the actionr line of the force is called the centre of percussion. Thus in Fig. 120, if a is the fixed axis, h will be the centre of percus- sion. It is also evident that, if ^ be a fixed object, and it be struck by the body AC^ rotating about a, the axis will not receive an impulse. AXIS OF INSTANTANEOUS ROTATION. 135. An axis through the centre of the mass, parallel to the- axis of spontaneous rotation, is called the axis of instantaneous rotation. A free body rotates about this axis. In regard to the spontaneous axis, we consider that as fixed in space for the instant ; but at the same time the body really rotates about the instantaneous axis which moves in space with the body. EXAMPLES UNDER THE PRECEDING EQUATIONS CONTINUED. 9. A horizontal uniform disc is free to revolve about a ver- tical axis through its centre, A man walks around on the outer edge ; required the ang>dar distance passed over by the man and disc when he has walked once around tlie circumfer- ence. Let W = the weiglit of the man ; w = the weight of tlie disc; r = the radius of the disc ; Oj = the angular velocity of the man in reference to a fixed line ; o> = the angular velocity of the disc in reference to the same fixed line ; Q = the f orcje exCrted by the man against the disc ; The result will be the same whether the effort be exerted suddenly, or with a uniform acceleration, or irregularly. We will, therefore, treat it as if it were an impulse. The weighta are here used instead of the masses, for they are directly pro- tm.\ EXAMPLES. 211 portioi.al to oacb other, and it is more Dataral to speak of the weight of a man thau the mass of a man. We have __ moment of the impulse "~ 7no7nent of inertia 2TF = CDi. W If, in a unit of time the man arrives at the initial point of the disc, we have which, combined with the preceding equation, gives If IT = t/7, we have ©1 = fTT, for the angular space passed over by the man, and o) = Jtt, for the distance passed over by the disc. 10. In Fig, 115 let the force F be constant; required the nu?nber of complete turns which the body O will make about the axis DE in the time t. Let r = the radius of the circle passed over by /^; ri = the distance of the centre of the body from the axis of revolution ; ki = the principal radius of gyration of the body in refer- ence to a moment axis parallel to DE'y k = the radius of gyration of the body in reference to the axis DE; then, according to equation (123), ^ = r,» + ^*; 212 ROTATION OP n«wi and, according to equation (152), ol?6 moment of forces moment of inertia Fr df Multiply by dt and integrate, and we have dO Fr dt = -^7^.t. MB the constant being zero, for the initial quantities are zero. Multiplying again by dt^ we find Fr (9 = ^IMI^ ^; which is the angular space passed over in time t ; and the nnm- ber of complete rotations will be e_ __ Frf 11. If the body were a sphere 2 feet in diameter, weighing 100 pounds, the centre of which was 5 feet from the axis ; i^, a force of 25 pounds, acting at the end of a lever 8 feet long ; required the number of turns which it will make about the axis in 5 minutes. 12. If the data be the same as in the preceding example ; required the time necessary to make one complete turn about the axis. 13. Suppose that an indefinitely thin body, whose weight is TT, rests upon the rim of a horizon- tal pulley which is perfectly free to move. A string is wound around the pulley, and passes over another pulley and has a weight, P, attached to its lower end. Supposing that there is no resistance by the pulleys or the string, required the distance »»• »i passed over by P in time t. [185.] SOLID BODIES. ilS According to equation (152), we have Ft Pg TrTPo-(Tr+P)r' 9 from which it may be solved. (This ifl equivalent to applying the weight P directly to tihe weight IT, M in Pig. 10, and hence we have, according to equation (21), W->rF d'^s _ p g dl?~ * (Pd d?B but referring it to polar coordinates, we have r~ = — -, which substituted reduces the equation directly to that in the text.) 14-. A disc whose weight is W is free to revolve about a horizontal axis pass- ing through its centre and perpendicular to its plane. A cord is wound around its circumference and has a weight, P, attached to its lower end ; required the distance tlirough which P will descend in t seconds. We have Pio. 128 d^ - ^^y Wk,^ + Ft" from which Q may be found, and the space will be rO, (This may be solved by equation (21). The mass of the disc reduced to aa equivalent at the circumference will be — -^ , and that equation will become —{Pr ^—^ )-3Tj = -P; which, by changing to polar coordinates, may be reduced to the equation in the text.) 15. If, in the preceding example, the body were a sphere revolving about a horizontal axis, the diameter of the sphere being 16 inches, weight 500 pounds, moved by a weight of 100 pounds descending vertically, the cord passing around a groove in the sphere the diameter of which is one foot ; required the number of revolutions of the sphere in five seconds. 214 ROTATION OP [13^.i Fio. 123. 16. Two weights, P and TF, are suspended on two pulleys by means of cords, as shown in Fig. 123, the pulleys being attached to the same axis G. No resistance being allowed for the pulleys, axle, or cords ; required the circumstances of motion. We have d^6 _^ moment of the forces dfi mo7nent3 of inertia _ P.AO-W.BO ~" P{AGf + W{Baf+disGAC.h^ 4- disoBCJii^' in which disc A G, etc., are used for the weights of the discs Let the right-hand member be represented by My then we have ^^ TIT. e = lMe. ' 17. In the preceding example let the discs be of uniform density, AG =S inches, BG= S inches ; the weight of ^ 6^ = 6 pounds, of GB = 2 pounds, of P = 25 pounds, and of W = 60 pounds; if they start from rest, required the space passed over by P in 10 seconds, and the tension of the cords. 18. A homogeneous^ hollow cylinder rolls down an inclined plane hy the force of gravity ; required the time, A ;2^ The weight of the cylinder may be resolved into two components, one parallel to the plane, which impels the body down it, the other nor- mal, which induces fricti m. Fio. 124. The frictiou acts parallel to the plane and tends to prevent the movement down it, and is assumed to be sufiicient to prevent sliding. Let W = the weight of the cylinder ; i ■=. the inclination of the plane to the horizontal ; ir= TTcos i = the normal component; m. •=. the mass of a unit ; the altitude = 1 ; J|1M.] SOLID BODIES. WS ^ = the coefficient of friction ; T = ^N = the tangential component ; jT' = ir sin ^ = the component of the weight parallol to the plane ; 7*1 = the internal radius of the cylinder; r = the external radius; d = the angular space passed over bj the radios ; 8 •= ACy the space. This is a case of translation and rotation combined, and equations (157) give W d'^8 g dtr dW _ T.r _ 2gTr and from the problem 8 — re. Eliminating 8 and T from these equations, we get d^^O __ 2^7* sin i d¥ - Zr^~Vr}' Integrating and making the initial spaces zero, we have gr sin ^ V /y?** Rin «, ^r^ sin * K ri = 0;the cylinder will be solid, and V g sm t' and hence, the time is independent of the diameter of tlio cyiin'ler. 11 ^1 = r, the cylinder will be a thin annulus, and ▼ g sm t lionce, the time of descent will be V\ times as long ai 316 THE COMPOUND [185.1 when the cylinder is solid ; the weight being the same in both cases. If it slide down a smooth plane of the same slope, we have =v/: 2s_ g sm t which is less than either of the two preceding times. THE PENDULUM. X 19. Zet a body he suspended on a horison^ tal axis and moved by the force of gra/vity ^ required the circumstances of motion. We have d^d _ TnoTTient of forces df moment of inertia Fio. 126 ^^ in which h = Oa^ the distance from the axis of suspension to the centre of gravity a of the body ; W = the weight of the body ; 6 = bOa; and let ki = the principal radius of gyration ; then the preceding equation becomes d^e gh df ~ h^ + Jcl sin 6, This equation cannot be completely integrated in finite terms, but by developing sin 6 and neglecting all powers above the first, we find for a complete oscillation which gives the time in seconds when A, hi and g are given in feet. To find the length of a simple pendulum which wiU vibrate nS6, 137.] PENDULUM. 317 in the same time, we make equations (5), page 196, and (161) equal to one another, and have I = ^L+A* = Od, (162) Let ad = Jii^ then ^ — A = Ai = -7- ; .-. h\ = k^ (163) 136, Definitions. A body of any form oscillating about a fixed axis is called a compound jpendvlum, A material particle suspended by a string without weight, oscillating about a fixed axis, is called a simjple pendulum. The point d is called the centre of oscillation. It is the point at which, if a particle be placed and suspended from the axis 6^ by a string without weight, it will oscillate in the same time as the body Od, Or, it is the point at which, if the entire mass be concentrated, it will oscillate about the axis in the same time as when it is distributed. The point O, where the axis pierces the plane xy^ is called the ce7itre of suspension, 137. Results. The centres of oscillation and of percussion coincide. (See Article 134.) According to equation (163), the centres of oscillation and of suspension are convertible. According to the same equation the principal radius of gyra- tion is a mean proportional between the distances of the cen- tres of oscillation and of suspension from the centre of gravity. Equation (161) indicates a practical mode of determining the principal radius of gyration. To find it, let the body oscillate, and thus find T, then attach a pair of spring balances to the lower end and bring the body to a horizontal position, and find how much the scales indicate ; knowing which, the weight of the body and the distance between the point of attachment and the centre of suspension O, the value of h may easily be computed. The value of g being known, all the quantities in equation (161) become known except ^, which is readily found by a solution of the equation. Fio. 12«. 518 CAPTAIN KATER'S [188.1 Examples. 1, A prismatic bar oscillates about an axis passing through one end, and perpendicular to its length; required the length of an equivalent simple pendulum. 2. A homogeneous sphere is suspended from a point by means of a fine thread, find the length of a simple pendulum which will oscillate in the same time. 138. Captain Kater used the principle of the convertibility of the centres of suspension and ' W oscillation for determining the length of a simple seconds pendulum, and hence the acceleration due to gravity.— PA^7. Tram., 1S18. Let a body, furnished with a movable weight, be provided with a point of suspension O (figure not shown), and another point on which it may vibrate, fixed as nearly as can be esti- mated in the centre of oscillation 0, and in a line with the point of suspension and the centre of gravity. The oscillations of the body must now be observed when suspended from O and also when suspended from O, If the vibrations in each posi- tion should not be equal in equal times, they may readily be made so by shifting the movable weiglit. When this is done, the distance between the two points 6' and O is the length of the simple equivalent pendulum. Tlnis the length Z and the corresponding time T oi vibration will be found uninfluenced by any irregularity of density or figure. In these experiments, after numerous trials of spheres, etc., knife edges were pre- ferred as a means of support. At the centres of suspension and oscillation there were two triangular apei'tures to admit the knife edges on which the body rested while making its oscil- lations. Having thus the means of measuring the length Z with accuracy, it remains to determine the tinie T. This is effected by comparing the vibrations of the body with those of a clodL The time of a single vibration or of any small arbitrary number of vibrations cannot be observed directly, because this would require the fraction of a second of time, as shown by the cl'jck, *o be estimated either by the eye or ear. The vibrations of the fl38.J EXPERIMENTS. 219 bcKly may bo counted, and here there is no fraction to be esti- mated, but these vibrations will not probably fit in witli the oscillations of tlie clock penduhim, and the differences must be estimated. Tliis defect is overcome by " the method of coinci- dences." Supposing the time of vibration of the clock to bo a little less than that of the body, the pendulum of the clock will gain on the body, and at length at a certain vibration the two will for an instant coincide. The two pendulums will now be seen to separate, and after a time will again approach each other, when the same phenomenon will take place. If the two pendulums continue to vibrate with perfect uniformity, the number of oscillations of the pendulum of the clock in this in- terval will be an integer, and the number of oscillations of the body in the same interval will be less by one complete oscilla- tion than that of the pendulum of the clock. Hence by a fiimple proiX)rtion the time of a complete oscillation may be found. The coincidences were determined in the following manner: Certain marks made on the two pendulums were observed by a telescope at the lowest point of their arcs of vibration. The field of view was limited by a diaphragm to a narrow aperture across which the marks were seen to j)as8. At each succeeding vibration the clock pendulum follows the other more closely, and at last the clock-mark completely covers the other during their passage across the field of view of the telescoi)e. After a few vibrations it appears again preceding the other. The time of disappearance was generally considered as the time of coincidence of the vibrations, though in strictness the mean of the times of disappearance and reappearance ought to have been taken, but the error thus produced is very small. {Encyc. Met, Figure of the Earth.) In the experiments made in Ilartan coal-pit in 1S54, the Astronomer Royal used Kater'fl method of observing the pendulum. {PhiL 7Van«.,lS5G.) The value of T thus found will require several corrections. These are called " Reductions." If the centre of oscillation does not describe a cycloid, allowance must be made for the altera- tion of time depending on the arc described. This is called "the reduction to infinitely small arcs." If the point of sup- port be not absolutely fixed, another correction is required 220 A STANDARD OF LENGTH. (13a ] (Phil, Tfans., 1831). The effect of the buojancy and the resistance of the air must also be allowed for. This is the ''reduction to a vacuum." The length Z must also be cor- '^cted for changes of temperature. The time of an oscillation thus corrected enables us to find the value of gravity at the place of observation. A correction is now required to reduce this result to what it would have been at the level of the sea. The attraction of the intervening land must be allowed for by Dr. Young's rule {Phil. Trans.^ 1819). We thus obtain the force of gravity at the level of the sea, supposing all the land above this level were cut off and the sea constrained to keep its present level. As the sea would tend in such a case to change its level, further corrections are still necessary if we wish to reduce the result to the surface of that spheroid which most nearly represents the earth. (See Camh, Phil. Trans., vol. x.) There is another use to which the experimental determina- tion of the length of a simple equivalent pendulum may be applied. It has been adopted as a standard of length on account of being invariable and capable at any time of recov- ery. An Act of Parliament, 5 Geo. lY., defines the yard to contain thirty-six such parts, of which parts there are 39.1393 in the length of the pendulum vibrating seconds of mean time in the latitude of London, in vacuo, at the level of the sea, at temperature 62° F. The Commissioners, however, appointed to consider the mode of restoring the standards of weight and measure which were lost by fire in 1834, report that several elements of reduction of pendulum experiments are yet doubt- ful or erroneous, so that the results of a convertible pendulum are not so trustworthy as to serve for supplying a standard for length ; and they recommend a material standard, the distance, namely, between two marks on a certain bar of metal under given circumstances, in preference to any standard derived from measuring phenomena in nature. {Rejport, 1841.) All nations, practically, use this simple mode of determining the length of the standard of measure, that of placing two marks on a bar, and by a legal enactment declaring it to be a certain length. tl89, 140.] KLLIPTICITT OF THE EARTH. 221 139. Form of the Earth. The pendulum furnishes one of the best means for determining the form of the earth. Let a = the equatorial radius of the earth ; h = the semi-axis ; € = the ellipticity of the earth ; then a Let m = ratio of the centrifugal force at the equator to the force of gravity at the same place ; ^ = length of a second's pendulum at the equator ; 4o = the length of a second's pendulum at the poles ; then, from the MSoanique Cdleste, tome II., No. 34, we have The value of m is g-J-y. The formula for the length of the second's pendulum when the length at Paris is taken as unity, is I = 0.996823 + 0.00549745 sin^^, when ^is the latitude of the place. See Puissant's TraiiS ds GeodSsie, page 461. By this means it has been found that € is about ^|^. Bow- dich, in his translation of the Mecanique Celeste^ p. 485, re- marks, '^ It appears that the oblateness (e) does not differ much from ^^(^, and may possibly be a little more, though some results give a little less." 140. Torsion Pendulum. If an elastic bar, CD^ be fixed at one end, and at the other end have two weights, A^ and A^^ rigidly fixed to it by means of the cross arm, A^ A2, then if the arm be turned into the position J^iB^, the elastic resistance of the bar p^q DC will cause the weights to move back Ap- to A1A2, and by virtue of the energy of ' ,.'■' C"-^ the weights at that point, they will pass -gQ' \.jb ihsX position, and move on until their ' fio. i2t. motion is arrested by the action of the elastic resistance of the 222 VIBEATIONS OF A [111.] bar ; after which they will return to their former position, thus having a motion similar to that of the common pendulum. This arrangement is called a torsion pendidurn. The motion will be the same for one weight as for two, but when the bar J)C is vertical, the arms CA^ and CA^^ should equal each other, and the weight A^ equal A^. 141. To find the force necessary to twist the rod DG through a given angle. Let F= the force at A^ perpendicular to the arm GA^ ; a -^ GA^= GA,', l = UG; a = A^GBx ; / = the polar moment of inertia of a tran verse section of the bar DG\ and O — - the coefficient of the elastic resistance to torsion. The moment of the twisting force, F^ will be and the moment of the elastic resistance will be (see Hesistaiie of Materials, 2d Ed., p. 206), hence, we have ^'-v Fa = OIp ■■^=^'^ar The weights Ai and A2 are not involved in this problem. If the angle be measured from some fixed line making an angle </> with the neutral position of AiAz^ then instead of a we would have a^ — (f>^ and the last equation becomes if the force be reversea, it will twist the bar in the opposite [142. J TORSION PENDULUM. 22S direction, making an angle a^ with the fixed line of reference^ and we would have T - 7 ^* ~ (t,). Adding these equations gives 2 -J = -^ (oi - oa). 142, To find the tifne of an oscillation. The bar CI) having been twisted by moving the bar from it» normal position A^A^ into the position B^B^^^ and then left to itself, it is required to find the time of moving to the other extreme position B^B^. We will neglect the mass of the rod AiAi, and that of the bar DC\ and thus simplify the solution, and secure an approximate result. T^t /^ = the moment of inertia of one of the bodies, Ai, or A^ in reference to CD as an axie , ^ — a variable angle measured Jtiom the neutral posi- tion, AiA^ ; and, considering 7^ as a variable moment, producing the variable angle 6, we have from the second of equations (157), and the value of I^a from the first equation of the preceding Article, Multiply by dd and integrate, and obserring that for ^ = a the angular velocity is zero, we find d^ 01 , ^ ^ hence dO .,__ A^ JL. Integrating again gives lYlJ. . _^0 224 DENSITY OP ri48.1 which, between the limits of and a, gives which is the time of half an oscillation ; hence the time for a full oscillation, or the time of movement from B^ to i?8, will be which is the time required. The times are isochronous and in- dependent of the amplitude. The value of G may be eliminated by substituting its value taken from the last equation of the preceding article. Making the substitution, we find Fa from which I and /have also disappeared. From this we find A 143. To find the density of the earth. The plan of determining the density of the earth by means of a torsion rod was first suggested by the Rev. John Mitehel. He died before he was able to make the experiment, but the plan was executed by Mr. Cavendish, who published the result in the PhiL Trans, for 1798. Subsequent to 1837, Mr. Bailey, at the request of the Astronomical Society (England), made a new determination of the result. He made upwards of 2,000 experiments with balls of different weights and sizes, and sus- pended in a variety of ways, a full account of which is given in the Memoirs of the Astronomical Society, Vol. xiv. We give here only some of the more prominent features of the ex- periment. The torsion rod DC was very small, so that it could be easily twisted. Two small balls, A^, A2, were suspended from the tl43.] THE EABTH. 225 torsion rod by a light cross-bar. Two large balls, By^ E^^ were placed on a plank which turned about a point O directly under C, and the whole so arranged that the centres of gravity of the four balls were in the same horizontal plane. The apparatus Fio. 128. was inclosed in a small room so as to exclude currents of air, and the weights E^ and E2 were moved into the desired posi- tions from the outside of the room by means of mechanism ex- tending into the room. The weights E^ and E^ were first placed nearly at right angles with the rod A^ A2, when the latter would assume some •neutral position as Oa, The balls EiE^ were then brought quite near to the small ones, Ei, B^^ when the attraction of the former drew the latter from their neutral position, and they oscillated about some position of equilibrium as J^i ^2- The angle aCB^ was observed, and also the time of the oscillation about the position OB^. The balls were then changed to the position i^/^, making the angle a 67^ as nearly equal as possible to aCE^ ; but it was found that the line Ca did not always bisect the angle E^ CE:^^ but the mean of many readings was taken as the most probable value. The angle E^CF^ will be Oj — 02, given in the preced- ing Article. It is proved, by the law of attraction, that the attraction of a homogeneous sphere is the same as if its entire mass were con- centrated at the centre of the mass, and varies inversely us the square of the distance from the centre. Let M = the mass of one of the large balla ; m = '' « u a « « small balls; 15 DENSITY OF [143.|i D = the distance between their centres ; and /It = the attraction of a sphere whose mass is unity upon another unit when the distance between theii centres is unity ; tlien the force of attraction of the mass M upon m will be Mm and this is the value of Fin the last equation of the preceding Article. But there being two balls in this case, the moment of this attractive force will be ' ^ Mm which (by neglecting the attraction of the large ball and plank upon the rod 61^2, and of the plank upon the small balls), equals the second member of the last equation of the preceding Article. Hence, Mm ^-^^ = A(ai - «2)(^j Let E be the mass of the earth, R its radius, and g the force of gravity, then E 9 = t^-m* Eliminating />t, and making /^ = m (a' + f /^), we have f =(a» + |^)(a.-a,)2g-(5,y The density of the earth is thus reduced to the determination •f the Oi — og between its two positions of equilibrium when • In Bailey's experiiaents, the value used was 5' = M^l-2€ + (fm-e) cos^A I ; In which « is the ellipticity of the earth, m the ratio of the centrifugal force at the equator to equatorial gravity, and A. the latitude of the place. 1144.] THE EARTU. 22T under the action of the masses in their alternate positioiis, and the time T of oscillation of the torsion rod. To observe these, a small mirror was attached to the rod at Cy witli its plane nearly perpendicular to the rod. A scale was engraved on a vertical plate at a distance of 108 inches from the mirror, and the image of the scale formed by reflection on tlie inirror was viewed in a telescope placed just over the scale, in this way an angle of one or two seconds could be read. The final result was that the mean density of the earth is 6.674:7 times that of distilled water at its maximum density. 144. Problem. If the earth were a homogeneous sphere^ at what point in the radius must it he struck, and what momentum must it receive, that it shall have a velocity of translation of V and of rotation of mf Let M =■ the mass of the sphere, R = its radius, Tc^ = \^R = the principal i-adius of gyration. (See Example 4, page 174), and a = the distance from t lie centre to the point where the impulse is applied. TJie momentum must be wherever the blow is applied. The moment of an impulse being the same as the moment of the momentum, we have, ac- cording to equation (155), moment of the momentum to = —. r ; laoment of inertia _ MV.a ^ _ Va 828 ANGULAR VELOCITY. L145.J The angular velocity of the earth per hour, i« 27r ^ 24 ' and the linear velocity in the orbit is 68,000 miles per hour nearly ; •*• "" - 2,040,000 ^' Letting It = 4,000 miles, we have a = 24: miles nearly. 145, PfiOBLEM. A homogeneous disc has a motion of tra/iM' lotion and of rotation entirely in its own^plane, when suddenhf <my point in the disc becomes fixed ; required the angular velo^ dty about the fixed point. Let F'= the velocity of translation of the centre of the disc; CD = the angular velocity about the centre ; p = the perpendicular distance between the fixed point and the line of motion of the centre at the instant that the point becomes fixed ; r = the distance between the fixed point and the cei.tre of the disc ; ki = the principal radius of gyration ; and ©1= the angular velocity about the fixed point. Then _ moment of the momentum, ~~ m,om,ent of inertia _ h^co+ Vp If F= 0, we have 1146.] OF FINITE MASSES. 229 If the centre becomes fixed, we have j? = 0, and r = 0, and (I). 1 46. Pkoblem. a sphere whose radius is R has an angvlaf velocity o)^ and gradually contracts until its radius is r ; re- quired thejlnal angular velocity. We will assume that the body remains homogeneous through- out, and that there is no change of temperature, and that the change of volume is due simply to the mutual attraction of the particles for each other, which is supposed to draw them towards the centre. Then will the moment of the momentum be constant Let 0)1 be the required angular velocity ; then we have from equation (155) Qa = fDM.^I^ = a)iJ/i.| r" ; .•.0)1 = ^01. Problem. — A spherical homogeneous mass w, radius r, contracts by the mutual attraction of its particles to a radius nr ; if the work thus expended be suddenly changed into heat, how many degfrees F. will the temperature of the mass be increased, its specific heat being * and the heat uniformly disseminated ? Consider the earth as a homogeneous sphere, mass Jf, radius i?, 5 its den- sity compared with water, g the acceleration on the surface of the earth due to gn^avity, g' that on the surface of ia. Then, since attractions are directly as the masses and inversely as the square of the radii, we have g' = j^* — r g. Initially, the force within the sphere varies directly as the distance p from the centre, but during contractions as the inverse squares; hence the acceleration at a distance a? of a particle originally at p will be iTt • '"^9 • ^. The mass of a spherical shell, radius p, is dm = (m -i- ^ irr^) 4wp''dp = -J p^dp. The force equals the mass into the acceleration, and this by dx will be an element of work ; hence the total work will be Dividing by J (Joule's mechanical equivalent of beat), also by the weight of a sphere of water equal in volume to that of the given sphere, or -^ . -— , JjL and also by the specific heat of the body, we finally have T — V-y-y- • — Problem by the Author in Mathematical Vufitor, July, 1880. CHAPTER XI, OBNERAL EQUATIONS OF MOTION. 147. P'Alembert's Pbinciple. — A body is a collection of material particles held together by a force exerted by each parti- cle upon the others. Having deduced the laws of action for forces acting upon a single particle, the direct process for de- termining the effect of forces upon a body of finite size, appears to be to consider all the forces which act upon each particle separately, including the mutual actions and reactions of the par- ticles, thus establishing equations for each particle of the body, and then to eliminate the terms involving the actions and reac- tions. But the latter are generally unknown. Various expedients were resorted to by the ancient mathematicians to reach the re- sulting equations, but the principle announced by D'Alembert greatlj' simplified the operations, and in many cases reduced the establishment of the equations to Statical principles. See WhewelPs History of the Inductive Sciences, Yol. 1., p. 365. The forces which produce the motion of a body may be applied at only a few points, and yet produce motion in every particle of it. These forces are called impressed forces. If we consider the particles as separated from each other, and forces applied to them which will produce the same motion that they had when in the body, the latter forces are called effective forces. The effective forces will then produce the same effect upon the body as the impressed forces, D'Alembert's principle consists in this, that if a system, of forces, equal and ojpjposite to the EFFECTIVE FORCES, act upon a body, they will he in equili- hrium with the impressed forces. In this principle no assumption has been made in regard to the character of the mutual actions and reactions between the particles, and hence it is applicable to flexible bodies and fluids, as well as to solids. It is equivalent to assuming that the forces T14a] GENERAL EQUATIONS OF MOTION. 231 within a body constitute a system which are in equilibriom among themselves. 148, To FIND THE GENERAL EQUATIONS OF MOTION OF A BODY. Let a?!, yi, Sj, be the coordinates of a particle whose mass is lUi ; J^i, J^i, Zi, the impressed forces parallel to the respective axes acting upon the particle ; and a similar notation for the other particles. The measure of the accelerating force parallel to the axis of X will be and if a force equal and opposite to this act upon the particle there will be equilibrium ; hence we have X,-m,-^^,=0- r,-m,-J = 0; Z, - m, ^ = 0. Similarly, /j*nf> (ill (JyZa and similar expressions for all the other particles of the body. But if 5'X, 2* Y, SZy be the sum of the respective axial com- ponents of the impressed J'o7'ces, then SX = Xi + Xg + Xg + etc. ; and similarly for the others. Ilence, if m be the mass of any .particle whose coordinates are a?, y, 2, at the time ty we have 4iccording to D'Alembert's principle and similarly for the others. Taking the moments of the forces in reference to the axis oi 05, we have, in precisely the same way, the equation and similarly for the other axis. Ilence, we have the follow ing six equations :~ 282 GENERAL EQUATIONS OF [148.1 ST ^"^'O XZ — Sm (164) - (166) I^t «!, yi, ^, be the coordinates of any particle of the body referred to a movable system whose origin remains at the centre of the mass, and whose axes are parallel to the fixed axes, and X, y, i, the coordinates of the centre of the mass referred to the fixed origin ; then we have X = X + Xi] Smx = Smx 4- HmXi ; But the origin of the movable system being at the centre of the mass, we have, from equations (71«) or (84«), Xmxi = 0; ttnd the last of the preceding equations becoraet ^m df ^ dj^x [148.J MOTION. 233 since -^- is a oommon factor to all the particles 97»; bnl and similarly for the others. Hence, equations (164) become (leu) Similarly, in equations (165), we have d?z d\ dJ^ d^Zi But y and i are common factors in their respective terms, therefore the expression becomes but, hence, we finally have 5^^' = 0; and similarly for other terms. In this way the first of equar tions (165) becomes ^Zy. - 27ei+l^ -SFe -2 0^4S-^SH ^34 GENERAL EQUATIONS OF fi4ai Multiply the third of equations (166) by y, the second by s, subtract the latter from the former, and we have My df Mz df :szy-srs: which, substituted in the preceding equation, gives Dropping S before X, Y, Z, and letting those letters repre- sent the total axial components upon the e?itire body, and Zyi — T%, etc., the resultant moments of the applied forces, we have the six following equations : TIT^ '7 (167) (168) Equations (167) do not contain the coordinates of the point of /application of the forces, hence, the motion of translation of the centre of a mass is independent of the point of applica- tion of the force or forces ; or, in other words, it is indepen- dent of the rotation of the mass. Equations (168) do not contain the coordinates of the centre of the mass, and being the equations for rotation, show that the rotation of a mass is independent of the translation of its centre. ^49, 150.] MOTION. 235 These equations are sufficient for determining all the circutL.- stances of motion of a free solid. In their further use the dashes and subscripts will be omitted. 149. If X, Y^ Z, are zero, we have ^d}x dx /d^x dx ^ ^ and similarly for the others. Transposing, squaring, adding and extracting the square root, give Id^ -^ dy^ -\- d^ , « = V ^ VC^^ 6V + 67»; (169) which, being constant, shows that the motion of the centre of the mass is rectilinear and uniform. This is tJie general principle of the oonskbvation of thb CENTRE OF GRAVITY. CONSERVATION OF AREAS. 150. The expression, ady — ydxy is, according to Article 112, twice the sectoral area passed over by the radius vector of the body in au instant of time. Hence, if ^{rnxdy — mydx) = dAi ; differentiating, we find ^/ d^y d^x\ d^Ai d'^x If there are no accelerating forces m -^ = ; and similarly for the others ; hence d^ ~ ' de "" ' ~df ~"' .\A^ — Cit\ A2 = c^\ As — c^t; (170) 16 236 CONSERVATION OP [151.; the initial values being zero. These are the projections on the coordinate planes of the areas swept over by the radius vector of the body. They establish the principle of the conservation OF AREAS. That is, In any system of hodies, moving without accelerating forces and having only mutual actions wpon each other ^ the projections on any plane of the areas swejpt over hy the radius vector art jproportional to the times. CONSEKVATION OF ENERGY. 251. Multiply equations (167) by dx, dy, dz, respectively, „(Id and integrate, and we have if^2 - Mv^ = 2f{Xdx + Tdy + Zds) ; and for a system of bodies, we have iSiMv") - iS{Mv,^) = SfiXdx + Tdy + Zdz). (171) The second member is integrable when the forces are directed towards fixed centres and is a function of the distances between them. Let a, h, c be the coordinates of one centre, »!, bi, Ci, of another, etc. ; X, y, z, the coordinates of the particle m ; r, /*!, etc., be the distances of the particle from the res- pective centres; I^, jp[, etc., be the forces directed towards the respective centres ; then, resolving the forces parallel to the axes, we have X= i^cos a f i^ cos a + etc. _« — X —,ai — X = F -\-Fi- +etc.; r r Y == ir^-^Zl + ir.h^ + etc.; r r Z == f'-^ + F^^^^ + etc. (151.J ENERQY. Multiplying by dx, dy^ dz^ respectively, and adding, we have dx + ^dy + 'Iz \ T T V I \ ri 7*1 ri + etc. But r» = (a-ajf + (*-y)» + (o-3)»; (172) and by differentiating, we find a — x^ h — y^ G — Zj d/r=— dx ^y dz ; fiimilarly, dri= - dx - - — -dy - dz ; n n n etc., etc., etc. These substituted above, give Xdx + Ydy + Zdz = — Fdr — F^dr^ — 7^2 - et<5. Therefore, if F^ etc., is a function of /*, etc., and /a, /^i, etc., the intensities of the respective forces at a distance unity from the respective centres, or F=fi<l>(r); Ft = fi,(f>{r,); * etc., etc. ; the second member, and hence the first, will be integrable. In nature, if a particle m attracts a particle mi, the particle mi will attract m, each being a centre of force in reference to the other, and both centres will be movable in reference to a fixed origin. But one centre may be considered fixed in ref- erence to the other, and consequently the proposition remains true for this case. The second member of equation (171) being integrated be- tween the limits a?, y, z and aj^, y^, z^ we have \:^ {Mv^- mMvo^) = 2m<p {x, y, z) -2mcp {x,, y,,z,). (173) Hence, the gain or loss of energy of a system, subject to forces directed towards fixed centres and which are functions of the distances from, those centres^ is indej^endent of the j>ath 338 CONSERVATION OF [151. K desiyi*ihed hy the bodies^ and depends only upon the position left and ari'ived at hy the bodies^ and the intensities of the forces at a unites distance from the respective centres. Therefore, when the system returns to the initial position, or to a condition equivalent to the original one, the vis viva will be the same. From equation (171) we have The gain or loss of energy in passing from one position, to another equals the work done hy the impressed forces. Let Wq — the work done by the impressed forces in passing from some definite point (a?o, 2/o> ^o?) to an- other definite point (a?i, ?/i, 2i,) ; and W — tlie work done by the same forces in passing from tlie first point to any point (a?, y, 2,) ; then from equation (ITI), we have i2{Mv,') - i:^{Mv') = W: and subtracting the latter from the former, gives l2{Mv') - i:S{Mvf) = TFo - W ; t>r, by transposing, in which the second member is constant. The first term of the first member is the kinetic energy which the system has at any point of the path, and the second term is t>he work which has been done by the forces upon the body and has become latent, or potential ; hence, in such a system the sum of the Jdnetic and potential energies is corir slant. . This is the principle of the conservation of energy in theoretical mechanics. This term has been extended so as to include the principle of the transmutation of energy as estab- lished by physical science. If a portion of the universe, as the Solar System for in- Btance, be separated from all external forces, the sum of the kinetic and potential energies will remain constant, so that if the kinetic energy diminishes, the potential increases and the 1151.J EXEROY. 23^ converse. If external forces act, the potential and kinetic energies may both be increased. To be more specific, suppose that the earth and sun consti- tute the system, the sun being considered the centre of the force. The velocity of the earth will be greatest when nearest the sun, and will diminish as it recedes from it. While reced- ing, the amount of work done against the attractive force of the sun will be ^Mi^ - iMv^ = - IF; in which J/" is the mass of the earth, v^ the maximum velocity, and V the velocity at any point. Tlie second member is nega- tive because the first member is. When the earth is approaching the sun the velocity is in- creased and the living force is restored, and the kinetic added to the potential energy is constant. Again, if a body whose weight is W be raised a height A, the work which has been done to raise it to that point is Why and in that position its potential energy is Wh. If it falls freely through that height it will acquire a velocity ^ = V 2^A, .-. Wh=zYv' = iMiP ^9 which is the kinetic energy. If the same body fall through a portion of the height, say ^1, its kinetic energy will be WJi^ = ^^"^it ^^^ the work which IS still due to its position is W {h—h-^^ which, at that instant, is inert or potential. It is found, however, that in the use of machines or other devices, by which work is transmitted from one body to another, all the work stored in a moving body cannot be utilized. Thus, in the impact of non-clastic bodies there is always a loss of living force. (See Article 32.) This, so far as theoretical mechanics is concerned, is a loss, and is treated as such, and until modern Physical Science established the correlation of forces it was supposed to be entirely lost. But we know that in the case of impact heat is developed, and Joule determined a definite relation between tho quantity of heat and the work necessary to produce it, and called the result the mechanical equivalent of heat. Further 340 ROTARY MOTIONS. (!«,] investigations show that in every case of a supposed loss of energy, it may be accounted for in a general way by the appearance of energy in some other form. It is impossible to trace the transformation of energy as it appears in mechanical action, friction, heat, light, electricity, megnetism, etc., and prove by direct measurements that the sum total at every instant and with every transformation remains rigidly con- stant; but by means of careful observations and measurements nearly all the energy in a variety of cases has been traced from one mode of action to another, and the small fraction which was apparently lost could be accounted for by the imperfec- tions of tlie apparatus, or in some other satisfactory manner; until at last the principle of the conservation of energy is recognized to be a law as universal as that of the law of gravi- tation. The exact nature of molecular energy which manifests itself in heat, chemical affinity, etc., are unknown, but, accord- ng to the general law, all energy whether molecular or of Anite masses, is either hinetia or potential, COMPOSITION OF ROTARY MOTIONS. 152. Take the origin of coordinates at the fixed point when there is one, or at any point of the axis of rotation when the body is free. Conceive three cones having a com- mon vertex at the origin of co- ^ ordinates and each tangent re- .'^TNI spectively to the coordinate planes; J I ^ then will the angular velocity of I that element of the cone which for j the instant is in contact with the 7! ^ plane xyj considered as rotating about the axis of 2, be defined as the angular velocity of the hody about that axis ; and similarly for the other axes. Let &?„ o^y^ od^^ be the angular velocities of the hody about the respective axes x, y, z. If ea, considered as infinitesimal, be the actual velocity of a particle in th© plane xy, positive from x towards y^ 9— Oc^ ca = pGDg\ them cd — poD^^ cos (180° — acd) — — gd^ • p sin JlOc = — oDgyf da = pGDg sin (180° — acd) = oo^ - p cos XOe — oo^^ r fl52.J ' ROTARY MOTION. 241 Similarly in regard to the axis of y, we have •nd in regard to x^ When all these rotations take place at the same time, we have, by adding the corresponding velocities, the several velo- cities along the axes dt = OyS — ft)«y ; dy {1U\ dz dt = ©a-y — ©yOJ. The particles on the instantaneous axis have no velocity in reference to the movable origin, hence dx di = 0, dy_ dt = 0, ^^ -0. .-. G)y3 — (o»y = 0, (OjIX — coxS = 0, co^y — (OyX = ; (175) which are the equations of a straight line through the origin, a,nd are the equations of the instantaneous axis. Let a, ^, 7, be the angles which it makes with the axes x, y, 3, respectively. then (Anal. Geom.), cos a = &)ar VwJ' (O. (Oy ; cos B = ^ , - ~ a « ; cos 7 = o>« Vq);^^ + O)/ + Q),2 To determine the angular velocity of the body, take any point in a plane perpendicular to the instantaneous axis. Let the point be on the axis of x, and from it erect a perpendicular to the instantaneous axis, and we have xsma = xVl- C08*a 242 ROTARY MOTIONS. [153, 154.1 For this point y = and s = in equations (174), and we find for the actual velocity, F = dt — X yw^ + w? ; and her ce = - = V^i + < + <; (176) which represents the diagonal of a rectangular parallelopipe- don, of which the sides are ©a., q)^, w^. 153. Moments of rotation of the centre of the mass about the fixed axes. Multiply the second of equations (167) by 2, the third by y, subtract the former from the latter, and we have Mi dJ^y l.d'z ^dHj\ „ ^_ Treating the equations two and two in this manner, dropping the dashes, and substituting Zi, J/i, iV^i, for the second mem- bers, we have ,^/ d}x d}z\ (177) These equations are of the Bsuneform as equations (168). MOTION OF A BODY DURING IMPACT. 1 54, Motion of the centre of the masses. The second mem- bers of equations (167) are the accelerating forces. If any two of the bodies collide, they being free in other respects, ^the action of one body upon the other is equal and opposite to that of the latter upon the former ; hence, in regard to the system they neutralize each otlier, and the motion of the centre of the masses will be unaffected by the collision. If there are no (165, 156.] CONSTRAINED MOTION. 243 accelerating forces the velocity of the centre will be aniform and in a strai^^ht line, as shown in Article 140. To find the velocity of the bodies after impact requires a knowledge of their physical constitution. See Articles 28 and 29. 155. The motion of rotation of the centre of the entire mass about the origin will also be unaffected by the collision, when the bodies are acted upon by accelerating forces ; for, the mo- ments of the forces due to the collision will neutralize each other, and the second members of equations (177) will contain only the applied forces. This would be illustrated by the impact jof two asteroids, or in the bursting of a pi-imary planet. But the rotation produced in each body about the centre of its mass depends upon the moments of the forces applied to the body, and hence, upon the moment of the momentum produced by the impact. CONSTRAINED MOTION. 156. General equations of rotation about a fixed jpoint. Take the origin of coordinates at the fixed point. For this case equations (164) vanish. Substitute in (165), the values of -T^, etc., obtained from (174), and reduce. We have ^x d(Oy dcog and similarly for the others. Let Z, Mf iV, be substituted for the last term respectively of equations (165), and substituting the above values in the last of these equations, we find '=J!r. (178) 244 PBINCIPAL [IS?.! The other two equations may be treated iu the same manner. But they are too complicated to be of use. Since the position of the axes is arbitrary, let them be so chosen that 5'ma^ = 0, Smxz = 0, Smys = ; (179) in which case the axes are called principal axes ; and we will show in the next article, that, for every point of a body, there are at least three principal axes, eacli of which is perpendicular to the plane of the other two. Let a?i, yij %, be the principal axes, having the same origin as the fixed axes, and A = ^'in{y^ -{- Zi), the moment of inertia of the body about aJi; B = 2in{zi+iCi)^ moment about y^ ; C = 2m{x^+y^)y moment about % ; also let G)i, 0)2, 0)3, be the angular velocities about the respective axes osi, yi, and %, and substituting these several quantities in (178), we have (180) These are called Euler's Equations. The origin of coordinates may be taken at the centre of the mass, and as the rotation about that point is the same whether that point be at rest or in motion, as shown at the bottom of page 234, equations (180) are applicable to the rotation of a (free body when acted upon by forces in any manner. PRINCIPAL AXES. 157. At every point of a hody there are at least three prm cipal axes perpendicular to each other. When three axes meeting at a point in a body are perpen dicular to each other, and so taken that Smxy = 0, Smyz = 0, Smzx = ; ihey are called Principal Axes. 1167.] AXES. 245 The planes containing the principal axes are called Principal Planes. The moments of inertia in reference to the principal axes at any point are called the Principal Moments of Inertia for that point. Let OJS'he any line drawn through the origin, making angles a, y9, 7, with the respective coordinate axes. First find the moment of inertia about the line 01^. From any point of the line 01^, erect a perpendicular, NP, The coordi- nates of P will be X, y, z. Hence we have ^iV= ajcosa + ycosyS -f SC0S7; ^ 1 = cos'^a + cos^yS + cosV ^'°- ^•^• The moment of inertia of the body in reference to ON^ will be /= 2m. PN* = 2m {OP' - OIP) = 2m •{ («* + 2^» + 2' — (a? cos o + y cos iS 4- 2 cos 7)' } sslm -{ (aj*+y' + 2^) (cos^a + cos^jS + 008*7)— (« cos o+y cos 3+2 cos 7)' y = 2m {y^ 4- 2*) cos'^a + 2m {x^ + 2-) cos'-)3 + 2m (a;^ + y^) co8"7 — 22wy2 cos )3 cos 7 — 22m2a; cos 7 cos a — 22mxy cos a cos j8 =^co8'o+J?co8*/3+ Cco8''7— 22)cos /3 008 7—2^008 7008 0—27^^008 a cos /9 ; in which A,jB, 6^, have the values given on page 244, and D^E^F^ are written for the corresponding factors of the preceding equation. This may be illustrated geometrically. Conceive a radius vector, r, to move about in space in such a maimer that for all angles a, y9, 7, corresponding to those of the line ON^ the square of the length shall be inversely proportional to the moment of inertia of the body. Then in which c is a constant. Hence, the polar equation of the locus is -, -srA oyB*a+BooB*$-\' Ocoe*y-'2D cos /3 cos 7— 3^ cos y cos o— 2i^cos a cos f» S46 PRINCIPAL ri5ai Multiplying by /*, we have which is the equation of the locus referred to rectangular co5r- dinates, and is a quadric. Since -4,j5, and 6^ are essentially positive, it is the equation of an ellipsoid, and is called the momental ellipsoid. Therefore, the moment of inertia about every line which passes through any point of a body may be represented by the radius vector of a certain ellipsoid. But every ellipsoid has at least three principal diameters, hence every material system has, at every point of it, at least three principal axes. If the ellipsoid be referred to its principal diameters the coefficients of yZy sXyXy, vanish, and the equation of the ellipsoid becomes e = Ax^-hBf+ C^. In many cases the principal diameters may be determined by inspection. Thus, in a sphere every diameter is a principal axis. In an ellipsoid the three axes are principal axes. In all surfaces of revolution, the axis of revolution is a principal axis, ftnd any two lines perpendicular to each other and to the axis of revolution are the other two principal axes. 158. If a hody revolve about one of the princijjal axespasa- ing through the centre of gravity of the hody, that axis wiU suffer no strain fi'om tJie centrifugal force. Let s be a principal axis, about which the body rotates. The centrifugal force of any particle will be which, resolved parallel to x and y, gives mco^Xj mxo^y ; and the moments of these forces about the axis of z are, f Dr the whole body, Ztruo^xy, ^ma^yx ; 1 159, 160.T AXES. 247 but these, according to equations (179), are zero. If the body be free and revolves about this axis it will continue to revolve about it. For this reason it is called an axis of perTnaneiit no- tation. If the body he free^ and the initial rotation be not about a principal axis, the centrifugal force will cause the iiutantane- ous axis to change constantly, and it will never rotate about the jpermanent axis. If, therefore, we observe that a free body revolves about an axis for a short time, we infer that it revolved about it from the beginning of tlie motion. RELATION BETWEEN THE AXES 0?, y, 2, FIXED IN SPACE AND THB PRINCIPAL AXES a?i, 2/i, ^i, FIXED IN THE BODY. 159, If the body be free, take the origin of coordinates Oy Fig. 129, at the centre of the mass ; but if there be a fixed point about which rotation is forced to take place, take the origin O at that point. Conceive a sphere, radius unity, hav- ing its centre also at 0, The line PP' will be the intersec- tion of the planes xy and x^yi, and P one of the points where it pierces the surface of the sphere. Positive angles will be determined in substantially the same manner as positive moments* described in Article 54, page 80; thus, positive rotation will be from + x towards + y, from + y towards 4- 2, and from + z towards + a?; and the opposite directions will be negative. In the following figure the rota- tions are all positive, and the angles or amounts of rotation are represented as less than 90°. In passing froui one system of rectangular axes to another, * It may be observed that the relations of the axes x, y, and 2, in the follow- ing figures are not the same as on the preceding pages. Thus, for instance, on pages 7t) and 111 the axis of y is vertically upward, while in the following figures z will be in that position. While the author prefei-s the former arrange- ment, lor the reason that the axes x and y would then retain the same rela- live position in the printed page as is most common when only two axes arc used ; yet for the sake of conforming with the usage of most writers, and for .greater ease in comparing results, we have concluded to make this change. It is proper to observe that this will cause no change whatever in the preceding analysis, provided the ord^er of the letters be observed, to the exclusion of right and left handed rotation. 248 EELATIONS BETWEEN [169.1 the origin being the same, we may proceed as follows : The system may be turned in a positive direction about z as an axis, bringing OX to the positioii of OP ; then rotating it positively about OP as an axis, bringing OZ into the position OZx ; and finally a positive rotation of the system about OZ,iy bringing OP into the position OX ^ the final position oi OY being OY^, Fig. 129. Then let B = ZOZi, being the angle between the axes b and 2t which is also the angle between the planes xi/ and Xi yi ; ip = XOP, being the angle between the axis of x and the line OP; ^ (p = POXx, being the angle between the line OP and the new axis of a?i. In astronomical language the line OP is called the line of nodes, and PX^ the right ascension. Expressing the above steps in the process analytically, we have Euler's method of passing from one system of axes to the other. Thus, let x\ y\ z\ be the position of the new axes at the end of the first step, then we have {Coordinate Geometry y Art. 54, or Art. 219) [l».l AXES. X = x' COS Ip — y' sin 0,, y = x sin t\) + y cos v-, z = 2 . 249 Next, turn the system in a positive direction about a?' as an axis, through an angle ^, and let x\ y\ z", be the position of the axes at the end of the second step, then X = X , y' = y" cos 6 — z" sin ^, z' — y" sin 6 + z" cos 6. T-«r Fio. 130. Fio. 131 Finally, turn the system in a positive direction about s' through an angle ^, and let jTi, ?/i, ^i, be the final position of the axes, then x" = Xi cos <p — yi sin 9?, y" = a?i sin qj + yi cos <^, 2" = 2i. Eliminating a?', y\ z\ x'\ y'\ 2", be- tween these equations gives X = (cos cp cos ^ — sin ^ sin rp cos ^) x^ + (— sin ^ cos ip — cos ^ sin ip cos ^ yj + sin ^ sin 6 . z^ y = (cos q) sin tp + sin <^ cos ip cos ^ Xi + (— sin <p sin ^ + cos (p cos ip cos ^ yi — cos ^ sin 0. Zi 2 = sin <^ sin ^ . a?i + cos ^ sin 6^ . yi + cos ^ . Zi (181) 250 RELATION BETWEEN (160] We have, for passing from a system of rectangular axes to another system having the same origin {Coordinate Geometry, Art. 219), X = xi cos {xx^ + 2/i cos {xy^ + z^ cos {xz^ \ y = x^ cos {2jx^) + 7/1 cos (yy^ + z^ cos {yz^) V • (182) z = x^ cos (2,ri) + 2/1 cos ( «7/i) + z^^ cos (2^1) ) A comparison of equations (181) and (182) gives : cos {xx^ = cos (p COB tp — sin ^ sin ip cos ) cos (.r?/i) = — sin ^ cos ip — cos ^ sin ^ cos ^ I . (183) cos {xz^) = sin 6 sin ^ J cos (yx-i) = cos ^ sin ^ + sin ^ cos tp cos 6 ] cos (3/yi) =: — sin <^ sin tp + cos (p cos ^ cos 6 I . (184) cos (?/2;i) = — sin ^ cos tp \ cos (;2:ci) = sin 9^ sin 6 \ cos (zy^) = cos ^ sin 6 y . (185) cos ( zz^ = cos ^ J We Iiere have nine direction cosines all expressed in terms of 6, ip, (p. These may vary while the axes .Tj, yi, z^ remain fixed. The same relations may be found by tl)e solution of spherical triangles formed by arcs of great circles joining the points where the axes pierce the surface of the sphere before referred to. IQO, -Relation hetweeii the angular velocities of the hody about its respective principcd axes fixed in the hody^ and the angxdar velocities about the lines OZ, OZi^ OP, respectively, } In Fig. 129, -^ will be the angular velocity of the body at about OZ. —r- the anovular velocitv about OZu and -r-, thean- dt ^ *^ dt ' gular velocity about OP. For the })urpose of representing these quantities in the simplest manner, and for greater con- venience in the following analvsis, let the value of -^, the an- cLt pea] AXES. 261 gular velocity of the system about OZ^ be represented by a definite line laid off from 6>, positively (in this case) on the line OZ. Similarly, lay off a line of proportionate length on the line OZi to represent the angular velocity about that axis; and similarly on the line OP. This representation accords with a similar representation of statical moments as shown in Article 171 of the author's Elementary Mechanics^ and is also in accordance with the remark following equation (176) of this work. In a similar manner oo^ will be represented by a line on a?!, G72 on yi, oo^ on Zi. Tiiis being done, the angular veloc- ities may be referred to as lines; and the composition and resolution of angular velocities be treated in the same manner as the composition and resolution of forces, Articles 55 and 83. Hence, any one of the angular velocities will equal the sum of the projections on that axis of all the other angular velocities. Hence, by the aid of equations (185), we have : GTi = ^ cos ZOX^ + ^ cos Z,OX, dt de dt + ^COS POX^ + QO^ cos YxOX^ + G73 COS ZyOX^ d(p de — jT cos (zx^ + -^f- cos 90° + ^ COS a> dt ^ dt dt + 67;, COS 90° + G-3 COS 90° dip . . . de = -J- sin G> sin ^ + j7 COS cp dt dt ^ drh dq) w, = -,f eoszor, + "^ cos z,or, dt de + — COS PO Yi+ Ci7i COS XiO 1\ + GOs COS ZiO Yi lit dp . . de , = -J- COS <p sin e dt dt sin q) OOz ^ COS Z6>Zi + ^ cos ^6^Zi dt de + -J- cos POZi + GOi COS XiOZi + (Wj cos Yi02^ dtp ^ dcp = 41'"'^^ dt (186) 252 SPONTANEOUS AXIS. [161.] In a similar manner the values of -7-, -^7, -^. may be found, dV dV dt^ ^ ' but they may be as readily found by elimination. Eliminat- ing among (186) gives : de -— ~GJ. cos O) at GO2 sm (p dr/y _ GDi sin (p + GO2 cos q) dt "~ sin 6 -— = (»3 — cot d (cfi?! sin q) ■\- 00^ cos qy) (187) AXIS OF SPONTANEOUS ROTATION. 161. Considering the body as perfectly free, we have, ac- cording to the notation immediately following equations (165): a? = cc + a?!, (188) from which we readily deduce : and similarly, dx di ~ dx dx^ ' dt "^ dt dy dt ~ dy dy^ dt "^ dt dz di = dz dzi dt ^ dt (188) X being a coordinate referred to a system of coordinates fixed in space, x refers to the centre of the mass in the same sys- tem of coordinates, but a?i is a coordinate to the same partiole as a?, referred to a parallel system having its origin continually at the centre of the mass. Suppositions may be made arbi- trarily upon any one of the three terms in equations (188), and a corresponding relation determined between the remaining 1161.] SPONTANEOUS AXIS. 253 terms. Thus, i^ ^ = ^> ^^^^^ similarly for -^ and -^ ; ws have dx dx ,- which shows that the velocity of every particle is the same as that of the centre of the mass. This is as it should be, since there will be no rotation. and similarly for the others ; hence the velocity of any parti- cle in reference to the fixed origin will be the same as that in reference to the centre of the mass. This is evidently as it should be, since the centre of the mass will, according to the hypothesis, be at rest. Finally, if dx ^ dt/ ^ dz ^ nQl\ e have dt ~^' dt -^' dt -"' dx dt~ dxi dy dyi dz dzx dt' dt~ ~ dt' dt ~ dt (192) Generalizing equations (191) by multiplying by ?, m, n, re- spectively, and integrating, we have Ix = a, my = J, nz ^=^ g\ (193) where a, 6, c, are arbitrary constants of integration. Adding* we have he + 7ny = a -!r h^ nz + my = c + h; (194) which are the equations to a right line in space, and according to (191), this line will be at rest at the instant, in reference to the fixed origin. This line is called the axis of spontaneous rotation. 254' SPONTANEOUS AXIS. [162.] Since sd, y^, ^i, equations (192), refer to tlie same particles, it follows from these equations that the axial velocities of the particles on the axis of spontaneous rotation, in reference to the centre of the mass, are equal but directly opposite to the axial velocity of the centre of the mass in reference to the fixed origin, and hence, more briefly : The velocity of the axis of spontaneous rotation in reference to the centre of the mass is equal hut in opj^osite direction to that of the centre of the mass in reference to a fixed origin, D 162. To illustrate this in a simple ^y^ case, let AB represent a body whose /^i .^^—^ b' centre G moves in the line CC and A ^'^ IQ B which rotates about its centre. If ^ ED represent the position consec- utive to AB^ they will intersect in some point as a, then will a line through a perpendicular to the plane of the two positions AB and ED^ be the axis of Sfpontam,eous rotation. Conceive the line AB to move to the consecutive parallel position A! B\ CC representing the velocity of the centre; then turn the line about C as a centre, that point a which has the same velocity backward that (7 had forward will be a point in the axis of spontaneous rotation. If the body be a disc moving in the plane of the paper and having a uniform rotation about its centre, the spontaneous axis will have a uniform motion parallel to the line CG\ The combined motions of translation of the entire body and of rotation about the centre of the mass may be considered as a simple instantaneous rotation about the spontaneous axis. It will also be observed that the angular velocity of tlie body about its centre will, at the instant, be the same as that of the centre about the axis of spontaneous rotation (equations (192)). QuEBiES. 1. Is the spontaneous axis always within the body ? 2. In Fig. 133, when the bar AB falls into the line of mo- tion (7(7', where will the axis of spontaneous rotation be ? 3. If the centre of the line AB has a uniform velocity in a straight line, and at the same time the line has a uniform [168.J SPONTANEOUS AXIS. 255 angular velocity about its middle point, required the path of the extremity B, 4. If the line AB has a uniform velocity of 20 feet per second, and a rotary velocity of 10 turns per second, required the distance from the centre to the axis of spontaneous rotation. 6. If the uniform velocity of the centre of a disc be v feet per second, and the uniform angular velocity in the plane of the disc and of motion be oo ; required the distance of the spontaneous axis from the centre of the body. Let X be the required distance ; then XQO •=. v\ V .•.« = —. GO If r be the radius of the disc, x will be <, =, or > than r, according as v is <, =, or > than rco. Only one line, equa- tions (194), fulfils the condition of a spontaneous axis. RELATIONS BETWEEN THE SPONTANEOUS AXIS OF ROTATION, THE CENTRAL AXIS AND THE LINE OF ACTIO]^ OF THE RESULTANT. 163. Observing that a?, y, s, equations (174), are coordinates in reference to the centre of a free body as an origin (the sub- scripts having been dropped), and hence are the same as a?i, yi, Si, in equations (192), we have, by substituting the values of the former in the latter, dy dz (195) which are the equations to the axis of spontaneoys rotation^ the origin of coordinates being at the centre of the mass. The 256 RELATIONS BETWEEN THE [163.] equations to the axis of instanto/iieous rotation or central axis^ in the same notation are, equations (175), (restoring the sub- scripts), COyXi = GJx^yi J (196) A comparison of equations (195) and (196) shows that these lines are parallel ; hence, The spontaneous axis of rotation is parallel to the instanta- neons {or central) axis. Letting Y be the velocity of the centre of the mass, a, h, c, the angles between the line of motion and the axes x, y, ^, re- spectively, we have J- z= V COS a, -^ = V COS 6, -jr = K cos c. (197) Substituting these in (195), we find for the distance be* tween the central axis (196) and the axis of spontaneous rota, tion (195), {Coordinate Geometry, Appendix III.), omitting numeral subscripts. -fy {Gax-\-GOy) cos^ a +2goxGOz cos acosc + (goI + go'z) cos^ c* -r, 5 -^^ . (198) GDx + GOy + GOg Let the axis of y be the axis of instantaneous rotation, then will GJx =: Oy GJz= 0, and we have, dropping the subscript, h, = ^, (199) GO * This may also be written ^ _ V / (ttJ^ + &)\) COS^ a + 2ft3a, (»y cos « cos 5 + (ft3 J + GOx 1/ Go^) cos ^ b (00^ + 00^) Gos^b + 2G0y Go^ cos & cose + {goI -h <»|) cos* e _ y GOI + GOl + OOl [163.] SPONTANEOUS AND INSTANTANEOUS AXES. 257 a result readily deduced from Fig. 133, since uLtimatdy we would have CC = V= Ca. oj; hence Ca = hi = V -i- co. If the angular rotation and velocity of the centre both be uniform, hi (198) will be constant ; and it is also evident that the linear and rotary velocities may vary proportionately in such a way as to make hi constant. This is readily seen in the more simple case of equation (199), and illustrated by ex- amples 3 and 4, p. 206. If any number of forces, 7^, 7^, i^, etc., act upon a body producing both a translation of the centre of tlie mass and ro- tation about that centre, they will be equivalent to a single force Id applied at some point of the body ; for we have (Eqs. (85), (87), and (86)), ^i^cos « = X = 7? cos a 1 :^i<'cos ^ = Y = B co&h 2Fcos y — Z = E C0& G ^ /. E = VX^ + F* + Z^ ; (200) (201) from which Jd becomes known, and hence «, J, c, equations (200), also become known. To find the line of action of E\ since the moments of the separate forces are known, 2F cos y . t/i — 2F cos fi .Zi, = L (say) becomes known, and similarly for the others ; hence Zyi - Yzi = L ^ Xzi - Zxi = M Yx,-Xyi^N] (202) where aji, yi, ^i, are the coordinates of the line of action of E in reference to the centre of the mass as an origin ; hence (202) are the equatknis to the line of action of the resultant. The third of (202) is a consequence of the other two. Since any point in the line of action of a force may be taken as its point of application, the point (xi, yi, z^) may be considered as 17 258 SPONTANEOUS AND [163.} the point of application of the resultant. To find where the re- sultant pierces the plane y^, make a?i = in (202), and we have, iV M and similarly for the other planes. If the forces reduce to a couple, we have i? = 0, and there will be no motion of the centre produced by this syetem of forces, although the centre may have a motion due to initial conditions. In this case the left members of (202) all reduce to zero, and the point of application will be found to be infi- nitely distant. The right meir^^^i's for statical equilibrium w^ould also be zero, but if Z, J/, iV, one or all, have finite values, they will produce rotation, and must be placed equal to the left members of (168). The inclination, 6^ between the line of the resultant (202) and the central axis (196), will be {^Coordinate Geometry^ Eq. (6), Art. 199), dropping the numerical subscripts, cos B = ^ (204) V(X^ + r^ + Z'O (G.1, +(4 + 00l) If i? = 0, Equation (201), cos ^ = tt? which is indetermi- nate. If the forces are parallel to the axis of x, then T' = 0, Z = 0,003. = Oy and cos (9 = 0; that is, the action line of the resultant will be perpendicular to the central axis, and hence also perpendicular to the axis of spontaneous rotation. The shortest distance between the ac- tion line of the force, equations (202), and the spontaneous axis, equations (195) and (197), will be (Appendix III., Coor- dinate Geometr?/), dropping the numerical subscripts, , . (^^f/-^) (". -In) - (I -.- ^--) (^- 5^) ^.... [163.] INSTANTANEOUS AXES. 259 As a special case, let there be a single force (or forces having a single resultant) acting parallel to the axis of a? and at such a point as to produce rotation about the axis of y only, and let this be a principal axis, then we liave : X= 7?, J^ = 0, j^r .= 0, Z = 0, ^^ = 0, 1> and (205) becomes, omitting subscripts, 90°, B Fig. 134 and (204) gives , If V It 00 cos ^ = ; .-. d= 90°, (206) hence the spontaneous axis will be perpendicular to the action line of the force, will lie in the plane yz^ and be at a distance from the action line equal to the value given by equation (206). The first three conditions immediately preceding equation (206), in equations (167) give, after integrating once, E being considered constant. m-j- =mV= lit at dy tt^'^ dt = ^2 (206a) where tw, the mass, is used to distinguish it from Jf, the mo- ment, in (206) above. The velocities Cj and c^ are constant, and are unaffected by the force or impulse 7?. If ("i and r^ are not zero, "Fwill not be the actual velocity of the centre, but will be the velocity produced by the resultant 7?. From the first of (206a) and (199) we have -, 7)1 V tnhxoo (207) 260 SPONTANEOUS AXIS. [MlJ In equations (180) we will have and, considering the moment of the forces as constant, we have by integrating '-^^^M; (208) and hence (206) becomes ^ = 1' + ^'^ = '^' « In Fig. 134, h being the point of action of the force in reference to the body AB, and a the projection of the axis of spontaneous rotation, then I = ah, hi = Oa. Let Ob = Ag, then from (209) we have Z-/., .= /,, = -^; ,'.7iA=h'', (210) and since k^ is constant, it follows that in the plane containing the action line of the force and the centre of the mass, the spontaneous axis and point of ajjplicatimi of the force are con- vertible. In other words, if a be the point where the spon- taneous axis pierces the principal plane xz when the force is applied at b in the same plane, then if the force be applied at a the spontaneous axis will pass through b. (See also Prob- lem 8, page 209). Although, in establishing equation (206), we have as- sumed a constant force applied at some point in one of the principal planes of the body, yet these conditions are not always practicable. Examples 4 and 5, page 206, illustrate the case when the forces are constant and parallel. Even in these cases it is necessary also to assume that the body has no initial velocity, or if it has it must be the same as might have [163.] MOMENTS OF MOMENTUM. 261 been produced by the action of the force, or forces, under the conditions above imposed. But all these equations are applicable to the case of instan- taneous forces, or, in other words, of impact, the initial veloci- ties either being zero or abstracted from the above conditions. The equation to the line passing through the origin of coordinates — which is at the centre of the mass — and the point where the action line pierces the plane yz^ equations (203), will be found by dividing one equation of (203) by the other, and is — restoring the subscripts — y. = - -^^. (211) Let 9 be the angle between this line and the axis of spon- taneous rotation, and we have (195), (211), (See Appendix III. of Coordinate Geometry)^ cos & — (212) For the case of impact this may be reduced to a form which will lead to an interesting general result. Since this condi- tion excludes accelerating forces, JT = 0, etc., in (167) and (168), and integrating once, we have d3c dy dz M''^^ a,Mj=C,,M'f=C,', (213) dt dt f dz. dvA J. [T'^^di -'"'^-di)=^' dyA _ dx dz^ / dXx dz{\ y'^'^dt -'"''^-di) ^fwari^ - 'myi dxA dt) ^1 r > (214) where Zj, J/i, iTi are the moments of the momentum, as shown by the left members of the equations. Substituting in (214), 262 ROTATION. [168.1 fo^ -^ etc., their values from (174) (since (174) are equally true for the origin at the centre of the mass), and making the axes principal ones by the conditions of equations (179), we finally have A GOx. = ^ 3L M, ."^^1 ~ ^m {z\ + xt)' B aHz. iV, (215) Substituting in (212) the values of <Bx,, oa,^, ooy^, from (215) gives cos B' = _ __. (216) A f^\ 1 k / (^ A\' (B N^ + 1. If, in (216), j& = 0, cos ^' = ; that is if two of the prin- cipal moments of inertia equal each other, the spontaneous axis of rotation will be perpendicular to the line joining the centre of the mass and the point where the action line pierces the plane of those moment axes. If the line of impact be parallel to the axis of x, Zi, equa- tions (215), will be zero, which gives go^^ = 0, and since -^, and-^ will also be zero, these in (195) show that the axis of spontaneous rotation will be in the plane yz. If now B = O, the axis of spontaneous rotation will lie in the plane of the moment axes of B and C^ and be perpendicular to the line drawn from the centre of the mass normal to and intersecting the line of the impulse where it pierces the plane of these axes. Further, if the line of the impulse be in the plane xy, making any angle with those axes, then Ml = 0, and JVi = 0, and the [164.] CENTRE OF PERCUSSION. 263 denominator becomes infinite, and hence 0" = 90° ; hence the axis of spontaneous rotation will be perpendicular to the plane xy^ or parallel to the axis of z ; and since this result is inde- pendent of the angle which the line of the impulse makes with the axis of a?, it will be true when it is parallel to that axis, and in the plane ajy, in which case the spontaneous axis will lie in the plane yz and still be parallel to the axis of z. This investigation beginning at equation (211) may be still further generalized by drawing from the centre of the mass a line normal to and intersecting the line of the impulse and finding the angle 6" between this line and the axis of spon- taneous rotation, but the results so found will be but little, if any, more general than those given above. CENTRE OF PERCUSSION. 164. It appears from Article 162 and Fig. 133, that the point a in the axis of spontaneous rotation may be considered at rest at an instant, and hence if the elements on that axis were held rigidly in space when the body is struck at 5, Fig. 134, the axis would suffer no shock. Such an axis generally exists, as shown by equation (205), or more simply by equa- tions (206), (209) and (210). Any point in the line of im- pact is called th£ centre of jpercvssion in reference to the axis of spontaneous rotation, and the centre of percussion in ref- erence to any axis is any point which may receive a blow without imparting a shock to that axis. Equation (210) enables one to find the centre of percussion in reference to any assumed axis. In the case of a compound pendulum acted upon by gravity only, the resultant force passes through the centre of gravity of the oscillating body, and the centre of percussion is farther from the axis of suspension than is the centre of gravity. In the ballistic pendulum, the blow should be at the centre of percussion, if possible, to avoid shock upon the axis of suspen- sion. Queries. 1. Can the centre of percussion be at the centre of gravity of a body? 2. Can the axis of spontaneous rotation ever pass through the centre of gravity of the body ? 264 CONSEKVATION. [165, 166.) CONSEEVATION OF THE MOTION OF THE CENTRE OF THE MASS. 165. Any condition that will render the second members of (167) zero, gives which integrated gives dx ~di — ^1? dy ~dt 62, dl dt — ^8 J and X = C\t + B, (217) (218) (219) when 6\, etc., B^^ etc., are constants of integration. Eliminat- ing t gives B^_y B, B, t'. 64 6i (220) which are the equations to a right line. Equations (218) give the velocities along the axes, and are constant. If G be the resultant velocity, we have c= ^c\' + a,' + a,- (221) The second members of (187) will be zero when no forces are acting npon the system ; also when a system of forces all act towards the centre but have zero for a resultant; also when the forces are the mutual actions between the parts of the system, for the resultant of such forces is zero ; hence When a hody or system> of bodies is acted ujpon hy any sys- tem of forces having zero for a resultant^ the viotion {if any) of the centre of the mass will be rectilinear amd uniform, CONSERVATION OF THE MOMENT OF THE MOMENTUM. 166. Any condition which renders the second members of (168) equal to zero, will give, omitting the subscripts, lUW.) OP THE MOMENT OF MOMENTUM. 265 and simflarly for the other two equations. Integrating gives the three equations (214). Let the ac- tual path of any particle be projected in the coordinate planes, and consider that projection which is on the plane xy. Let r be the radius vector to any point of the projected path, 6 the angle between x and r^ then, (Fig. Ill), X = r cos 0^ y = r sin 6 : ,'. dx = — r sin Odd + cos Odr^ dy = /• cos ddd + sin ddr ; which substituted in the third of (214) gives :S^r^^=iri. (222). rffi But -37 = G7 is the angular velocity of the particle m about the axis of z, rco its velocity measured in a circular arc, mrco its circular momentum, and r.mroo the moment of the momen- tum, hence 'Smr^Go is the entire moment of the circular momen- tum. But this is the same as the moment of the momentum in the direction of motion. For if O be the origin, OA = r, AOC = dd, Aj), d tangent, Op perpendicular to the tangent from the origin, then will the actual moment of the momen- tum be where ds = AC. But ds - ^.^s^ B^ But < \.7 AC \ AB '. \ OA '. Op', Fia.lSg. ,.AG=ds= -^^.; 266 INVARIABLE PLANE. (167.] which substituted in the preceding expression gives and taking the sum of all the particles gives the left member of (222). But the right member, iTj, is constant, being the constant of integration. The same result may be shown for each of the other coordinate planes; hence the actual mo- ment of the momentum about the central axis will be constant and equal to K= VLl + M\ + ]Sr\. (223) Hence^ when a hody or system of bodies is acted upon hy forces directed towards the centre of the mass / or hy mutual action^s between the particles ; or any system of forces in which their resultant moment is zero^ the motnent of the momentunrb is constant, and the moment of the momentum jprojected on any plane is also constant. Article 150 admits of the same generalization. THE INVARIABLE PLANE. 167. Consider that the moment of the momentum on the respective coordinate planes is represented by a line perpen- dicular to the plane and of proportionate length ; then will K^ since it is constant, equation (223), be normal to s^ fixed plane. If the axis of z be the axis of this plane, Zj and J/j will be zero, and iVi = K\ that is, the moment of the momentum on the plane xy will be a maximum. Hence, under the same general conditions as in the pre- ceding article, there is a fixed plane in reference to which the moment of the momentum is a maximum, and constant. The plane on which the projections of the moments of the mo- mentum are a maximum is called \hQ i7ivariahle jplane. In the solar system, knowing the positions and motions of the planets at any time, the position of the invariable plane may be found. (168.J ' ATTRACTION. 267 MUTUAL ACTION BETWEEN PARTICLES. 168. The mutual action between particles, or bodies, is of the nature of attraction, or of repulsion, or of both these forces. We will first consider attraction. Newton's law of universal gravitation is — Two particles attract each other with a stress directly propor- tional to the product of their inasses^ and inversely as the square of the distance between theiri. Thus if m and iii are two masses considered concentrated at mere points — in otlier words, the masses respectively of two particles — 77 the stress due to the mutual attraction of two units of mass, tlie one upon the other, at distance unity be- tween them, F \\\Q stress between the masses iii and m due to their mutual attractions when tlie distance between them is a?, then F=I1^^:, (224) where F^ m, and rn! are in the same units as IT. Takin<^ the origin of coordinates on the line joinini>: ^ ^ ^ the particles, a/ the abscissa of m', x" < — <x;^ — ?-« — ;— « > of ?/i, and X the distance between them at ^ time ^, then for m we have Y\a. 136. ^'^ = i.eosl80«=-/I^, (225) dtr or ^ = ^J (226) or d^x' m di? """^V' and for m, or also of —■ ccf' =^ x^ differentiating which, gives 268 ATTRACTION OF 6j^x' — d?x" = (Px, in which substituting (225) and (226). we have -— =-7Z-(m + m')-^. [169.] dt'^ Integrating, ^^ cM-i I . ,s (a — X df ax (227) (228) where a is the distance between the particles when their velocity is zero. Integrating again, t 211 (771 + m')J L (ax - x^fz + a cos-' (-)'"]. (229) In the preceding investigation the mass has been conceived to be concentrated in a mere point ; it wdll now be extended to that of a mass of finite dimensions. 169. To find the attraction of a homogeneous sphere upon a particle exterior to it. Let ABD be a spherical shell, centre C, P the position of an external particle. Conceive two consecutive radial lines drawn from P, cutting the shell in the points A and B^ and join P and C. Let 6 = AFC, y = perpen- dicular from A on PC, ds = an ^... at A,PA = r, PC = c, CA - CB = p, dfi — thickness of the shell, d = density, a = the radius of the sphere, and m the mass of the particle at P. The revolution of the semicircle about PC as an axis will generate the surface of the shell. The area of a section of the shell at A whose length is ds, will be Fig 137. element of lenofth of the circle at A, PA dp . ds. rie».] SPHERES. 269 Let this area be revolved about PC as an axis through an angle dcp ; it will generate a solid whose altitude is yd<p^ and the volume so generated will be ydcp .dp.dsy and its mass will be Sydcpdpds. The attractive stress between this element and the element at P will, equation (224), be proportional to m . 6ydq)df)d8 the component of which along the axis PC will hp ^ dpdsycosd . d<p and the attraction between the entire sphere and the particle will be ^^ ppp r^- dpdsycosed<p K •'o •'o ^ . =2.^,rr ^£^22?^ (230) Jo JO r* In order to integrate this, y, 6 and r must be found in terms of ds ; or, including cfe, all must be given in terms of a single variable. Drop the perpendicular CE—p^ Fig. 137, from the centre, (7, upon PB^ then jp =. G sin^; differentiating dp = GQosede\ (231) also from the triangle PC A, f* — 2^c cos^ + (? = f^\ 270 ATTRACTION OF [169,] and since p will be constant for any particular shell, we have for that shell, by differentiating, dr dd~~ TG sin^ r — G cos^ ' From the Theory of Curves, ds" = d7^ + 7^d^, which, combined with the two preceding equations, gives ds _ pr dd r — c (iO&0 ' and this with (231) gives ccos6ds = ^ ^ « r — Gcosff From the figure we have r — c cos^ = Vp^ ""i^^ also y _P or r c' ^ c These in (230) give 4:7rmS r r pdppdp ^ Jo Jo V^^=y' where the former result has been multiplied by 2, since the same perpendicular C£^ corresponds to two elements A and jB. Integrating gives and integrating again. ^Jo^''^ 169a.J SPHERES. 271 — —^ — = m X 'volume oj t/ie sphere x -j mass of the spJiere ^r»«.>x = m X -~^ — . (232) That is, the attraction between any homogeneous pphere and a particle exterior thereto varies directly as the mass of the sphere, and inversely as the square of the distance between the particle and the centre of the sphere. It is independent of the volume of the sphere^ and is the same as if the entire mass he considered as at the centre of the sphere. For the same reason, if the mass m be also a homogeneous sphere, it may be considered as concentrated at its centre of gravity; hence the attraction between any two spheres varies as the product of their masses conjointly and inversely as the square of the distance between their centres. This result is the same as that given by Newton in his PrindpiaJ*' 169a. Mass and stress are often referred to by the common na,inG pounds ; but w!ien necessiiry they are distinguished as pounds of mass, or pounds of farce. If two homogeneous spheres of equal size and known masses be placed at a known distance d, from each other, and under such circumstances that they are free to approach each other under their mutual attractions ; then if the equal oj.posite forces J^, applied to each sphere just sufficient to prevent their approach, be ac- curately measured in pounds by an accurate balance, we have from the preceding article and equation (224) .•./T=i^— ,. (233) m^ ^ But this method of finding 71 is impracticable, chiefly on ac- count of the difficulty of measuring the exceedingly small * Principia. B. 1. Prop. LXXVI. Cor. 3. 272 ATTRACTION OF SPHERES. [169a.l value of n which would result from manufactured spheres of manageable size. The usual method of finding 77 and one suffi- ciently accurate in practice, is to consider the earth as a homo- geneous sphere whose radius is the mean radius of the earth. The stress due to the attraction between the earth and any- body at its surface equals the weight of the body, ov m' x g = F, which in equation (224) gives n = ^g, (234) where H is the radius of the earth at the place where the body is weighed. The same result follows from equation (227) by neglecting the mass m of the small body when compared with that of the earth, and substituting g for the acceleration, and H for a?, the distance. Let B = 20,850,000 feet, the mean radius of the earth; E = the mass of the earth, being about 5^ times that of an equal volume of water, page 227 ; g = 32.16, the mean value of the acceleration of a free body on the earth, p. 144 of Elementary Me- chanics ; then from (234) we have 3 X 32i X 2 _ 67 , , ~ 4 X 11 X 3.1416 X 20,850,000 ~ 1,000,000,000 ^' if the density of the unit sphere be the average density of the earth and a foot radius. From this result 11 may be found for any assumed material, since the attractive stress will vary directly as the mass, for the same volume. Equation (227) thus becomes completely determined. The unit of mass for the solar system is sometimes taken as that of the earth con- sidered as 5-J times that of an equal volume of water. Equation (227) is for motion in the line of centres of the bodies. If X169b, 170.] LEAST ACTION. 273 their initial motions be not in this line, the bodies will describe orbits, as shown on pages 180 to 190. In the solar system the mass of the sun so far exceeds that of any one — or even all — the planets, that the latter are neglected in determining theoret- ically the character of their orbits. The value of 71 does not enter this part of the problem. Having proved that the orbits are ellipses, their nuignitude and position are found by deter- mining three points from observation. The problem of 'three bodies' subjected to mutual attrac- tions and liaving arbitrary initial motions, has attracted the attention of the most eminent mathematicians. In the case of the solar system, the mass of the sun is so great that its posi- tion is considered as stationary, and the problem is reduced chiefly to that of perturbations. For a discussion of this prob- lem see works on Theoretical Astronomy. 169b. Repulsive forces of the nature of electricity between two bodies are supposed to vary inversely as the square of the distance between them. Elastic forces resist the displacement of particles from their normal position, and vary directly as the amount of displace- ment, whether it be a resistance to tension or compiession. In this case it may be found that the orbit of a particle will be an ellipse. In regard to constrained motion of a particle on a curve or a surface in space, equations (167) may be reduced to those of (143). If the body be finite and rotation be involved, equa- tions (168) will also be necessary. All the equations of statics are also contained in (167) and (168) ))y considering accelerations as zero. The resulting for- mulas are contained in the preceding pages. PRINCIPLE OF LEAST ACTION. 170. Let m be the mass of a particle, v its velocity, ds its ])ath during the time dt^ then is the value of the definite in- tegral 'ods 18 274 LEAST CONSTRAINT. [170a.J deiined as the " action " of the particle in passing from the former position Si, to the latter s^. Since v = ds -r- dt, the above expression reduces to i ^^* v'dt, h The principal proposition following from this definition is : When a system of bodies is acted ujpon hy forces directed towards the common centre of their mass ; or subject only to th^ir mutual attractions ; or hy forces tending to fixed centres ; then in moving from a given position to another the sum of the actions of all the bodies is less than if they had been con- strained to follow any path different from the one actually described. The proposition in -this form is due to Lagrange. It might, with propriety, have been called ''the principle of least vis viva.''' It is not fruitful in the solution of problems; but the proposition once proved, leads to the general equations (167) and (16.S). This proposition shows that if a particle be con- strained to move upon a surface under the action of parallel forces the path will be a geodetic curve. gauss' theorem of least constraint. 170a. Jf a system of material particles be in motiouy under the action of accelerating forces., the sum of the prod^ nets of each 7:>a?•^^<?^^ and the square of the distance between its place at the end of time dt^ a7id the place which it would have had under the action of the given for ces, and in the same initial circumstances^ if it were free, is a minimum. This theorem of Gauss was first given in Crelle's Journal^ Vol. IV., 1829. It contains all the equations of motion ; and its chief interest consists in presenting the subject in this peculiar form. CHAPTER XII. MECHANICS OF FLUIDS. 171. Matter, in regard to its physical properties, is infinitely diversified. The physical condition is conceived to depend upon the relation between the attractive and repulsive forces existing between the particles of the body. Thus, if the attractive forces exceed the repulsive, the body is a solid, as iron, stone, etc. ; if these forces are equal, the substance is a Uqutd, as water, alcohol, etc. ; and if the repulsive forces ex- ceed the attractive, the substance is called gaseous, as air, hydrogen, etc. Solids are not all equally rigid. Thus, steel is vastly more rigid than jelly. As the repulsive forces in- crease in relation to the attractive ones, the bodies become more and more plastic, as glass, iron, lead, jelly, tar, molasses, etc., passing gradually, and it may be imperceptibly, from the hardest and most unyielding substance into liquids. If the particles of a liquid are not free to move among themselves, the substance is called viscoics, as molasses, vinegar, etc. Liquids also pass almost imperceptiMy into gases. There is no definite line dividing one of these classifications from the one to which it is more nearly allied, and since it is impossible to reduce the general formulas of mechanics so as to make them practi- cally useful for all the cases which may arise, regardless of the properties of the substance considered, we make arbitrary classifications, as indicated above, and in each class treat of ideal substances. The ideal substances, or bodies, are perfect in themselves, and may not have a single representative in nature ; still the results deduced on these hypotheses may rep- resent, with a more or less close approximation, what actually takes place. The more nearly the real conditions approximate to the ideal conditions, the more nearly will the equations represent operations or facts in nature. Discussions involving 275 276 LAWS OF PRESSURE. [172, 173, 174.J the imperfect conditions of bodies — such as viscosity, friction, elasticity, etc. — are often classed under Applied Mechanics, 172. Definitions. — The following are the typical con- ditions usually considered : A RIGID BODY, OY perfect solid, is one in which its particles are assumed to retain their relative positions under the action of forces. The body is assumed not to change its form. jL perfect fluid is a substance in which its particles are perfectly free to move among themselves. The property of viscosity is thus excluded. A PERFECT LIQUID is a perfect fluid in which the attractive and repulsive forces are equal. Water is usually taken as the type of such a substance. A TERFEOT GAS is a pcrfcct fluid in which the repulsive forces always exceed the attractive ones. Such a substance would expand indefinitely if not restrained. Air is usually taken as a type^ though hydrogen is a more perfect gas. , A heavy fluid \^ one in which its weight is considered. 173. It was formerly supposed that water was incompress- ible, while it was known that air could be easily compressed, and for this reason fluids were divided into coinpressible and incompressible, or elastic and non-elastic / the former of which were called gases and the latter liquids. Although it has long been known that liquids are compressible, yet since the com- pression will be very small for pressures to which they will ordinarily be subjected, we still consider a perfect liquid as incompressible. LAWS OF PRESSURE. 174. The pressure upon any particle of a perfect fluid at rest is equal in all directions, for if it were not, there would be a resultant pressure which would produce motion of tlie particle, since it is assumed to be perfectly free to move. The force here considered is finite. The weight of the particle acted upon by gravity is infinitesimal, and hence if [175, 176.] LAWS OF PRESSURE. 2t7 it be at rest the upward pressure against the particle must ex- ceed the downward by an infinitesimal amount, tlte amount being equal to the infinitesimal weight of the particle. 175. The pressui-e of a perfect fiidd at rest, upon the sur- face of t/te cesael cantainimj it, will be normal to that surface at every point of it. For otherwise there would be a tangential component, and this would produce motion, which is contrary to the hypothesis. This proposition is also true in regard to any surface exposed to fluid pressure ; hence, if a body be immersed in a fluid, the pressure upon it will be normal at every point of its surface. The discussion in regard to the stress in a fluid might be founded on Article D, p. 154. 176. Emry external pressure upon a perfect fluid at red wiU he transmitted with equal intensity to every part of tJie fluid in the vessel. For if there were any unequal pressures thus transmitted, motion would result. This is called The La/w of Ii^iuid Trans- mission. It is independent of the form of the vessel. The pressures due to the i<?e/</A^ of any particle of the fluid will also be transmitted equally to all parts of the fluid in the vessel helow the point where the particle is located. It cannot be transmitted above that point, for its weight is equilibrated by the upward pressure at that point. This proposition is illustrated by means of a vessel having closely fitting pistons at different parts jA of it, and the vessel filled with water, -^ when it is found that a pressure on I^ Ji any one of the pistons [the pressure ^^^^S-^^^v^:^ A being p pounds per square inch] ^^^^^S§^~~z~ ^^^^ produces the same pressure per square ^^^^^^^^g^ . inch on each of the other pistons. ^^^^^^^^^^^^F The pressure per unit area is called ^^^^^^^^^^S(^^ the intensity of the pressure. The ^^^Sm^^^^^K proposition is not rigidly proved by i'w- i3«. this experiment, for the pistons cannot work without friction. 278 LAWS OF PRESSURE. [177, 178.] 177. The pressure upon the base of a vertical prismatic vessel containing a perfect heavy fluid equals the weiglit of the ^mdijplus the pressure upon the upper surface of the fluid. For, according to the preceding article, the pressure upon the upper base is transmitted to the base with undiminished intensity ; and by the same article the entire pressure due to the weights of the particles is transmitted to the base, and must then be supported by the base. 178. The pressure upon the base of any vessel containing a heavy perfect fluid equals the weiglit of a prism of the fluid having for its base the base of the vessel, and for its altitude the height of the fluid ; jdus the pressure per unit area on the upper base into the area of the lower base. It is independent of the form of the vessel. For, if over any element of the base a vertical prism of the fluid be conceived of the same weight as that of the liquid, and the upper surface be subjected to a pressure of the same in- tensity as that of tlie given fluid, the press- ure on the element will equal the pressure on its upper base j^^'^s the weight of the vertical prism, Article 177. But this i c pressure will be transmitted equally in all .Fig 139. directions, Article 176, and hence pro- duces an equal pressure on every element of the base, and thus balance the real pressures. The intensity of the transmitted pressures is the same throughout the containing vessel, Article 176. If, therefore, the vessel be oblique, so that no real verti- cal prism can be erected on any element, the pressure will re- main the same, depending upon the vertical height. The ideal vertical prism being suppressed the proposition will be estab- lished. If /S' = the area of the base of the vessel, a = the depth of the fluid in tlie vessel, 6 =■ tlie weiglit of a unit of volume of the fluid, p = the pressure per unit of area upon the upper base of the fluid, P = the total pressure upon the base of the vessel ; -^^-= g__^ jrr^^ ^^^a ^ m^ {179, 180.] LAWS OF PRESSURE. 279 4;hen we have P = daS+pS= {Sa + p)S. (236) 179, Static Head. — In the preceding Article, conceive a prismatic vessel having the same base and filled with the same fluid to such a height, h, as to produce the pressure P upon the base, then P = ShS. (237) which compared with equation (236) gives h=a+^. (238) o This value of A is called the head due to the pressure, or the reduced head. It is a height of the fluid which would produce the actual pressure upon the base. The pressure varies directly as the head. In the case of gases confined in small vessels, the weight of the fluid, com|)ared with the external pressures, may generally be neglected, in which case we have 6=0, and (236) becomes F=pS, (239) Gases, confined in a vessel, are always subjected to a press- ure at the upper surface; but in the case of liquids, p, equa- tion (236), may be zero, in which case a becomes the same as h in (237), and we have /^ = §=f, (m where/?' is the pressure upon a unit of area of the base of the vessel. 180. Free Surface. — If the upper surface of a fluid is not subjected to a pressure, it is called SL^ree surface ; and is sometimes so called in the case of liquids when the atmosphere causes the only pressure. Liquids may have a free surface; but, according to the definition, a gas cannot have a free sur- face, since it may expand indefinitely. When the surface is 230 MECHANICS [181.1 free the pressure on the base of the vessel is given by equa- tion (237). 181. Pressure on a vSubmerged Surface. — The sub- merged surface may be the interior of the containing vessel,, or the surface of any body within the fluid, and of any form. The normal pressure upon any element of the surface will be, Articles 176 and 179, ^ = dxdA, where x is the reduced head, dA the element, and d the weight of a unit of volume of the fluid at the place of the element. The entire normal pressure upon the surface will be [ SxdA, (241) and if the fluid be incompressible, or if its density be uniforni> throughout the surface considered, 6 will be constant, and we have P=s[ xdA, ' (242) If ^ be the reduced head over the centre of gravity of the surface considered, then, equations (79), taking the origin in the surface of the fluid vertically over the centre of gravity of the surface, we have xA xdA ; .-. P = SxA, (248) that is : The normal pressure of a fluid against any stir- face submerged in it, equals the weight of a jprism of th^ fluid whose hase equals the area pressed and whose altitude is the reduced head over the centre of gravity of the area pressed. If the fluid be an incompressible liquid having a free sur- 1182, 183.J OF FLUIDS. 281 face, the reduced head will be the actual head over the centre of gravity of the area pressed. 182. Rksolved Pressures. — The pressure in any fixed di- rection will be the sum of the components of the normal press- ures in that direction. If 6 be the angle between the normal and required direction at any point of the surface, we have resolved pressure = 6 xdA cos 6. dA cos 6 is the projection of the element on a plane normal to the required direction, which call dB ; then resolved pressure = S \xdB = 6xB, (244) where x is the reduced head above the centre of gravity of the projected surface. Hence, when the projections of the ele- ments are not superimposed The component of pressure of a heavy perfect fluid upon any submerged surface hi any direction^ equals the weight of a prism of the fluid having a hase equal to the projection of the surface on a plane normal to the direction, cmd whose alti- tude is the reduced head above the centre of gravity of the pro- jected elements^ each considered to be at the depth of tlce corre- ponding surface elements. 183. Resultant Pressures. — If a body be submerged in a perfect, heavy fluid, the resultant of the horizontal pressures will be zero ; for the projection of all the ele- ments upon parallel planes will be equal, and hence the opposing pressures, according to the preceding article, will be equal, and their re- sultant zero. For the same reason the result- ant horizontal pressures upon the interior sur- face of such a vessel will also be zero. The resultant vertical pressures will also be zero, except that due to the weight of the displaced fluid, Article 176, hence a 282 MECHANICS [184.J the resultant pressure against a submerged surface will be upward and equal to the weight of the displaced fluid. 184. Centre of Pressure. — The point through which the resultant of all the pressures upon a surface passes, is called the centre of pressure. Let AB be a submerged surface, CD the line of inter- section of its plane with the free surface, or the upper surface of the fluid for a re- duced head. Take the origin at any point on the line (7i>, y along ED^ and x along EF. FG = the liead on the element dA = dxdy at F, Let 6 = FEG = the incli- nation of the surface pressed to the hori- PlG. 141. zontal, then GF = X &m 6 ; and the pressure on the element, w being the weight of a unit of volume, will be wdA . X sm 6, and its moment in reference to the line CD will be w dA . Q^ sin ^, and the moment for the entire surface will be w sin d ix^dA. Denoting the distance to the centre of gravity of the area^ by X, and the distance to the centre of pressure by l^ we have for the reduced head over the centre of gravity : X sin 0, and hence for the total pressure on the surface wAx sin 0, [184.J OP FLUIDS. 283 and for the moment of the pressure, wA^ sin 6.1; hence we have, by equating the preceding valneSi wAM sin^ = w sin^ o^dAl that is, Articles 137 and 134, TJie centre of pressure coincides with the centre of percussion^ the axis of rotation heing iii tlie free surface ; or what is the same, it is the moment of inertia of the surface divided by its statical moment. It is inde- pendent of the inclination of the plane, and of the density of the fluid. Examples. 1. Required the entire pressure upon the interior of a cons filled with water^ and standing on its hose. Let r = the radius of the base of the cone, and h = its altitude. The weight of a cubic foot of water is 62} pounds. The area of the base will be tt/^ ; hence the pressure upon the base will be 62i . 7rr\ L The normal pressure on the concave part will be 62i.2;rr.iV/''+ h\^h, or I X 02} X Ttrh Vr^ + h% which added to the preceding gives 884 MECHANICS [184.1 2. Required the normal pressure upon the interior of tlie cone in the preceding example when inverted. 3. Required the normal pressure upon the interior of a sphere filled with water, and compare the result with the weight of the water. 4. Find the normal pressure upon the interior of a cylindric- al vessel including its base, when filled with water. 5. Find the pressure upon the interior of a cone filled with water, the axis being horizontal ; the radius of the base being 1 foot and the altitude 4 feet. 6. Required the centre of pressure of a plane triangu- lar surface immersed in a fluid, the base being in the free surface. The moment of inertia of a triangle in reference to its base as an axis is, Article 104, Example 3, Its area will be i hd^ and the distance to its centre of gravity \d\ hence. Equation (245), we have 7 _ T2_^ _iJ ^- \hd? -^'^• 7. Required the centre of pressure of a rectangle hav- ing one end in the free surface, a being the breadth and d the depth. 8. Find the centre of pressure of a rectangle immersed ver- tically in a fluid, its upper end being a distance h and lower end d below the free surface, and a its breadth. 2 ^'-W. Ans. 3 d^-h^' 9. A cone standing on its base is filled with water ; required the vertical pressure upon the concave part, the radius of the base being r and the altitude h. 1185.] OF FLUIDS. 286 10. In Example 9 show that the pressure upon the base minus the vertical pressure upon the concave part equals the weight of the water. 11. Tlie concave surface of a cylinder filled with a liquid is divided by horizontal sections into ii annuli in such a manner that the pressure upon eachaniiulus equals the pressure on the base ; the radius of the base being /•, required the altitude and breadth of the mi\\ annulus. Ans. Depth h = nr. Breadth of 7)bi\\ annulus = '>/rh \/sJ m —^}n — IJ. 12. A rectangle, breadth 14 feet, depth 30 feet, is innnersed vertically in a liquid with one end in the free surface ; re- quired the distance below the free surface of a line which divides the pressures equally. Ans. 21.213 feet. 13. A vessel, in the form of a paraboloid of revolution, stand- ing on its base, is filled with water ; required the normal press- ure on the concave part, and the vertical upward pressure on the same, the radius of the base being IJ feet, and altitude 4 feet. Flotation. 185. Consider the case of a body in an incompressible liquid. Let Y be the volume of the body, I) its density ; V the volume of the liquid displaced, and 6 its density. Then, ac- cording to Article 183, the pressure vertically upward will be hence, if there be equilibrium, we have gDV=gSY\ or ^=:^; (246) 286 MECHANICS [186.1 that is, The volume of the hody will he to that of the displaced liquid as the density of the liquid is to that of the body. If then V= F'; lor the body will be entirely submerged, but if then or only a part of the body will be submerged, and the bo4y is said to float. The intersection of the plane of the free surface with the floating body is called the plane of flotation. The line joining the centre of gravity of the solid, G, and the centre of gravity G of the displaced liquid is called the axis of flotation, and if this line be vertical when the body is in equi- librium it is also called the line of rest. If the body be displaced from its line of rest, the vertical through the centre of gravity C of the dis- placed liquid is called the line of support ; and the point M where this line intersects the line of rest is called the 7netacentre. Fob the equilibrium of a floating body it is necessary that the line of support shall coincide with the line of rest, and the equilibrium will be staUe if the metacentre for an indefi- nitely small displacement is above the centre of gravity of the solid ; for in this case the reaction of the liquid along the line of support tends to turn the body toward the line of rest. ' If the centre of gravity of the body is below the centre of gravity of the displaced liquid, there will also be equilibrium. Fig. 142. 186. The depth of flotation may be found by means of [186.] OF FLUIDS. 287 equation (246) when the density and form of the body and density of the liquid are known. For example, to lind the depth of flotation of a paraboloid of revolution with the vertex downward : Let b = the radius of the base, h the altitude, and x = the depth of immersion. From the equation of the meridian section we have also, b* = 2ph ; The volume of the solid will be and of the displaced liquid, iTTi/'x; and these substituted in Equation (246) give Examples. 1. Find the depth of flotation of a solid sphere whose radius is 6 inches ahd density J that of tlie liquid in which it floats. 288 MECHANICS [187.1 2. Find the depth of flotation of a cone whose altitude is 5 feet, radius of the base 8 inches, and whose density is one-half that of the liquid in wliich it floats, the axis being vertical and apex upward. 3. Required the depth of flotation of a solid paraboloid of revolution, base downward, radius of 'base r, altitude A, and density | that of the liquid in which it floats. 4. Required the diameter of a spherical cavity in a uniform spherical shell of iron so that the depth of flotation shall be equal to the external radius of the shell, the external radius being r^ and density 7 times that of the liquid in which it is submerged. Ans, r = r 4 /_ 14* f: 6. Required the pressure necessary to just submerge a cu- bical block of wood each of whose edges is a feet, and whose density is i that of the liquid in which it is submerged. 6. In a uniform spherical shell, external radius r, density 7, required the radius of the cavity that the plane of flotation shall be tangent to the top of the cavity. Ans. r — 0.95'/' + . SPECIFIC GRAVITY. 187. If an external pressure act upon the body, either forc- ing it up or down, thereby producing equilibrium, we have, when the force F acts vertically down on the body, F^ijDY^gdY'', (247) for, the weight of the body, gBY, added to the downward pressure will equal the vertically upward pressure of the liquid. If the force Tracts upward, then the upward pressure of the liquid and the force F will equal the weight of the body; Iieuce gBY^gdY'^F. (248) T188.1 OF FLUIDS. 289 By means of these formulas, the weight of a body com- pared with the weight of an equal volume of the liquid may be determined. If the liquid used be selected as a standard, the relative weight thus found is called the specific weighty ov specific gravity. The body weighed in a vacuum gives directly, W==gDV', (249) then immersing it in the standard liquid and ascertaining the value of F necessary to produce equilibrium, we have from the preceding equations, W=gSV' ±F', W . . F or gdV 1 ± gd V ' (260) where -t- i^^is used when the body is heavier than the liquid, and — F when it is lighter. Water at a fixed temperature (usually 60° F.) and pressure (about 29.92 in. of the barometer) is usually taken as the standard. For a further development of the subject see the Author's Elementary Mechanics, 188. Wlien a mass of liquid is in motion under such con- ditions that its form becomes permanent, certain problems pertaining thereto may be solved by the principles of Statics, We notice the two following P RO BL EM S . 1. A heavy perfect liquid having a free surface^ is moved in a given direction tuith a con- stant acceleration y r-eqidred ^ the character of the free sur- faxte. Let the vessel move hori- zontally under the action of a constant force F, produc- ing an acceleration/, the weight of the liquid being TT— the weight of the vessel being neglected. 19 290 MECHANICS PSai Then we have, Equations (20) and (21), W=Mg, and F=M/; .-. -^=r = -^ = a constant. F f But F -^ W h the tangent of the angle which the result- ant of the forces of the free surface makes with the horizontal^ which, being constant, shows that the slope of the free sur- face is constant, and hence it is a plane. 2, tated A free^ heavy ^ perfect liqxdd in a cylindrical vessel is ro^ with a uniform velocity about its vertical axis / reqidred the form of the free surface. Since the forces will be the same in every meridian plane, we may consider the form in one meridian section, as xz for instance. The acceleration to which every particle is subjected is that of gravity, downwards, and the resistance to the centrifugal force radially in- ward. Let 00 be the constant angular velocity; then will the centrifugal force of a particle at a distance x from the axis of rotation be (Equa- Fig. 144. tion (142)), Z= - mg ; GD^X ' which is the tangent of the angle of the resultant force with, the horizontal : dx _ g ^ ' ' dz ~ OD^X ' [189.] OF FLUIDS. :2yi which integrated gives, the equation of the common parabola ; hence the free sur- face is that of a paraboloid of revolution with its axis ver- tical. Compare this result with that of Problem 8, page 195. A surface to which the resultant of all the forces at each and every point of it is normal, is called a le^jel surface. Questions. — 1. Will the parameters of all the paraboloids of the level surfaces at different depths be the same as that of the free surface ? 2. Will the intensity of the pressure at the circumference of the base of the vessel be the same as at the centre ? 3. If the revolution be so great as to cause the centre of the free surface to touch the base of the vessel, or even to expose a portion of the base, will the free surface still be that of a paraboloid of revolution ? Remark. — Since the paraboloidal surface is produced by a uniform rotation, and is raised from a free horizontal surface, it will be subject to oscillations. Could steady motion be con- tinued for a long time, these oscillations would very nearly disappear. It has been proposed by some writers to resort to this i)rinciple for the construction of very large concave mir- rors for astronomical purposes, but the delicate physical con- ditions, and the well-nigh jp^^<?c^ mechanism necessary for its success, are obstacles in the way of this undertaking. We are not aware that such a mode of making a mirror has been at- tempted. FLUID MOTION. 189. Definitions, — When the velocity in magnitude and direction at every point of a fluid vein is constant, the motion 292 MECHANICS [19a] is said to be " steady." In steady motion the path of any particle is called a " stream line." If throughout a finite portion of a fluid mass the motion of any element of that portion consists of a translation and a distortion only, the motion is said to be " irrotational " — a term used by Thomson and others. When the particles of a fluid have a " rotational " or " vor- tex " motion, a line drawn from point to point so that its di- rection is everywhere that of the instantaneous axis of rotation of the fluid, is called a " vortex-line." If through every point of a small closed curve we draw the corresponding vortex-line, a tube will be obtained called a " vortex-tube." The fluid contained within such a tube cun stitutes a " vortex." 190. Bernoulli's Theorem.— Let a particle, in steady mo- tion, trace the stream line AB ; and similarly another particle at G indefinitely near A, trace the stream line CD. Trace a small closed curve through the points A and C; then will all the stream lines through the elements of the curve Fig. 145. j^Q form the elements of an ideal tube, and may be replaced by an actual tube conceived to be destitute of friction ; and since there is steady motion the tube will be filled at all points, and constitute an elemenU ary .stream. The quantity of fluid passing any two points, as A and B^ in the same time, will be the same, and in this sense the flow is said to he permanent Let Si be the section of the tube at A^ S that at B, Vi the velocity of the flow at A^ v that at B ; then S,v, = Sv, which is the volume of the liquid flowing m a unit of time. Letj9i be the intensity of the pressure exerted at Ay and^ that at B^ then will j^i^i [190.] OF FLUIDS. 293 be the entire pressure on the section S^ and 2>iSii\ will be the work done by j?i in a unit of time, since the veloc* ity may be considered constant for an element of time, and may represent the space passed over in a unit of time. Sim- ilarly, the work done by p in the opposite direction in a unit of time will be Take any datum plane, above which are the ordinates 2^ to A and z to B, then the work done by gravity while the liquid is passed from the height Zi to that of 2, will be, w being the weight of a unit of volume, wSiVi{zi — z). The difference in the kinetic energies at A and B will be wSiVi 2^ (v'-vl). Since, according to the assumed condition, there is no re- sistance between A and B, the liquid between these points may be discarded — or, what is better, we may conceive that the liquid just before and behind the respective elements at A and B serve as a pair of perfectly flexible pistons which yield just enough to keep the tube full at all points passed by this elementary mass. Then will the entire work done upon this mass in passing from A to B equal the difference in the kinetic energies at those points, or jhS,v, -pSv + wS<o, {z, - s) = ^' (^ - «?) ; (251) or pAvt + wSiO,z^ + '^ v\ = pSv + wS,v,z + "^ v". (252) 294 Bernoulli's theorem. [loi.l In this equation, piS^v^ is the potential energy of the ini- tial pressure (Articles 26 and 151) ; wSiViZ^ is the initial po- tential energy due to gravity in reference to any arbitrarily assumed horizontal plane ; and ^ —^^i^^ is the initial kinetic energy of the mass w -^ g. The sum of these will be con- stant for steady motion. The second member of the equation represents corresponding quantities for any point of the stream ; hence For steady Tnotion without resistances, the sum of the jpoten- tial and Mnetic energies is constant This is BernoulWs Theorem. It may be expressed in another form, for dividing Equation (252") through by wS^v^ we have ^• + ,. + f = ^+, + f; (253) w 2g w 2^ where ^1 -r- w? is the head due to the initial pressure, z^ the ini- tial head in reference to the datum, v\ -^ 2g the head due to the initial kinetic energy ; and similar general expressions apply to the second member. The sum of the heads in each member is called the total head; hence For steady motion without resistances, the total head in ref- erence to any horizontal j^lane is constami. 191. Discussion.— If the extremities of a stream in steady motion be in the atmosphere, the pressure at the ends, j^i and j9, will be that of the atmosphere, and in most cases will be practically equal in magnitude, but opposite in direction ; for which condition (253) becomes ^ + ^=« + ^. (254) 2^ 1g The initial velocity is often so small that it may be neglected, for which case v^ = 0, and (254) becomes v^ = 2g{z^-z), (255) 1192.] TORRICELLl'S THEOREM. 295 and if the datum plane passes through the lower end of the stream, we have a = 0, and (255) becomes a/2^2i, (256) which is called TorricelWs Theorem. Comparing (255) or (256) with the lirst of (16), we have In steady motion vntliout resistances^ tJie head due to the velocity equals the height throntjh which a body must fall to acquire that velocity. If the datum plane passes through the lower point consid- ered, we have 2 = 0, and (254) gives V = Vv'i + 2y2i. (257) 192. To REPRESENT EQUATION (253) GRAPHICALLY, let AJ^ be a stream having a steady motion subject to different press- ures along its path, and as- sume AHt equal to Zi and draw the horizontal line Bill. Let AC represent the head due to the pressure at A, which may be that due to the atmosphere, or the at- mosphere and any other extraneous pressure, and CE the head due to the actual velocity at A. Then will AE be the total head above A, and if HiH be the datum, then will HiE be the total head above the datum. At ^we will have E a D '^s& ^ ■-=-^:^-^ ~^ ^s-S3^=;z:^ ^M -^ a ~~ ^^^^^^ F ^=^-^ ^^™ H i, Ft n. 14fi. f" \ z = HB; p-r'W=BF; v* -i- 2(/ = BF. If a vertical tube be inserted in the stream, having an open- ing up stream, the liquid should rise to the height ED; but if it be turned so as to be open sidewise, it would rise only to the height CF. The latter is called the hydraulic head. ^96 MECHANICS [1981 If the initial velocity be neglected, the horizontal line ED will pass through C. If at any point the pressure in the stream is zero, the line CF will be depressed and touch the stream at that point. Should it fall below the stream, the pressure would be nega- tive, but as liquids have no tensile strength, this condition would destroy the " steady " motion, and the equations would not be applicable. 193. If a vessel of varying sections be left free to discharge itself, or generally if a fluid has a " steady " flow through a pipe of varying sections, the pressure of the fluid in the small sections will be less than that due to the statical head, frictional re- sistances being abstracted. Let xS\ be the section at D, S that at -5, then vS = ViSi ; -vl = SI - S^ n-a *^l » Fig. 147. and (253) becomes w w (258) where z, — z = BD, — = head due to the pressure on I), = Aj w T (say), and ^ tlie liead due to the pressure at B, If Sy = xS^, we have w (259)* or the pressure will be exactly that due to the head. If S > Si, the last term of (258) will be positive, and henca [193.] OF FLUIDS. 297 jf>rr w will exceed the head given by (259), and we may write for this case ^ = Ai + BN. w Let the section at A be less than at D or aS < a^, then hence AM will be less than the height of the free surface above A, Also ^ = h, + AM, w If 8 is so much less than xSi that is negative, then j9 -j- w will be negative, and there will be a tendency to a vacuum. Let 6^ be such a section. Then if a bent tube CEG be inserted at 0, having its outer end below a liquid, the fluid from F will rise in the pipe a height FG^ 60 that w F0 = -^. (260) Examples. 1. A surface elementary stream of water having a velocity of 16 ft. per sec. undergoes chatiges in its sectional area as it passes a vessel which are proportional to the numbers 4, 6, 4, 3, 4, 6, 4 ; in what way can the head remain constant ? Draw a vertical contour of the stream with figured dimensions or distances. We have, since v'^ varies inversely as the square of the sec- tion, ^ OC 3V> -sVj V^5 h tV» 3-b> tV- 298 EXAMPLES. 1193.1 Since the stream is in the surface j? will be constant, being the pressure of the atmosphere, therefore z or the height must vary, and since the head remains absolutely the same the in- crease of z must be the same as the decrease of 7:;-. v = 16on entrance, and -^ ^-r^ — pd = ^ ^t. Hence, as the successive ^2 values of ^r- are ^ = 0,Y,0,-^,0,\SO; or the vertical contour will be of wave form. Fig. 148. 2. Water flows without loss of head through, a horizontal pipe of a diameter varying uniformly from 3 in. to 1 in. at smallest section, and then gradually enlarges. The velocity on entrance being 7 feet per second, what will be the mini- mum pressure at entrance, in order that the pipe may run full- and what may be the maximum diameter of exit into the at- mosphere ? Since the pipe is horizontal, z = Zi in Equation (253), and according to the conditions of the example, jp = at the smalls est section. We also have V at entrance = 7 feet ; /. V at smallest section = 7 x 9 = 63 ; . £1 49^(63)2. ••^ +2^- 2^^ ' .-.^1 = 26.44 lbs. For maximum diameter of exit into the air we have in the same equation p^ = 0^ p = 14.7 lbs. per square inch, Vi = 63, and V — 63 -rd^; (19a] EXAMPLES. 299 .-. d = 1.22 inches. 3. Water flows from a tank through a pipe, the lower end of which is 15 feet below the entrance, the sectional area of the pipe at the tank end being twice that of the lower end. Near the tank the pipe is perforated or broken ; And the head of water in the tank necessary to prevent air leaking in, or water out, through the fracture. Tiie pipe must be full, and the motion "steady." At the ends p = pi = 34 feet (or 14.7 lbs). If V be the velocity at the joint, then 2v will be that at exit, and we will have 34 + ^ + 15 = f^+ 34; .\ ^r-= 5 ft. = head in vessel. 2^ 4. Water is flowing from a reservoir through a siphon pipe, the discharge end of which is 20 feet below the level of the reservoir. The diameter of the pipe is 2 inches at the dis- charge end, and 2J inches at the highest point of the siphon. Neglecting all resistances, find the height to which the siphon may be raised above the reservoir. At ^,^1 = 0; at B, j} = 34 feet ; and Zi /^^N = 2 + 20 4 A, in (253). Equating heads at A and By Fio. 149. 800 MECHANICS [191f in which v is the velocity of exit. '=»4,i-(iyi- and Equating heads at O and B, ^ + 34 = 20 + 34; ,'. v^ = 40^, A=14 +20 {l-(gyj- = 25.Tft. If CB = hi we would find tnd ^ = ^h ; from which h^ may be found in terms of A. FINITE RESERVOIRS AND FINITE ORIFICES. 194. If the liquid in a vessel of finite size, Fig. 150, is free to run out through an orifice in the base — or side of the vessel, the course which the elements will take may be ob- served by introducing into the liquid some color- ing matter. In this way it is found tliat the fillets starting from the upper surface form curved paths which approach the orifice as a common point. In deflecting the paths of the particles, Fig. 150. centrifugal forces will be developed, from which it follows that the opposite sides of the fluid streams will be subjected to unequal pressures, which, however, in the case of perfect fluids, will not affect the flow, except as it changes the length of the patli. It is found, also, that the acceleration of the particles is different in the different elementary streams. 'I'liT'i/iifi// \^;!i'/;.'/,W [194.] OP FLUIDS. 301 As the streams approach each other at the orifice, they inter- fere, and by their mutual actions produce a contraction of the vein as it leaves the orifice ; the point of greatest contrac- tion being at a distance from the orifice equal to about one- half of its diameter. This view of the problem of flow leads to an extreme]}^ difficult, if not strictly impossible, analytical solution. We therefore adopt a more simple, and at the same time a sufficiently practical hypothesis, called the principle of the parallelism of sections, which implies that sections parallel before motion remain so during flow, and that equal volumes pass the parallel sections in equal times. If h be the height of the free surface of a liquid above the section of greatest contraction, we have from Equation (255) V = -v/2^A ; (261) or, 77ie velocity of dhcharge equals that of a hody falling in a vacuum from the free surface to tJie orifice. Experiments show that in some cases this result is nearly realized in practice, while in some extreme cases, depending upon the form and conditions of the orifice, it is twice too large. In practice, the several cases are classified as mere ori- fices, short tubes, reentrant tubes, etc., etc., and the velocity as determined by direct experiment in each of the cases, divided by the theoretical velocity, is called the coefficient (or moduhos) of velocity. Thus it is shown that when the discharge is through a thin plate, or past a well-defined sharp edge, the stream is at first contracted, forming the so-called vma con- iracta. The diameter of the section of greatest contraction is about 0.8 that of the orifice ; hence its section will be about 0.0-1: that of the orifice. For this case, the actual veloc- ity at the section of greatest contraction is ^^o- 1^^- found to be about 0.97 of the theoretical, and hence we would have V = 0.97 \/2^ (262) 302 VELOCITY OF DISCHARGE. P^l If the flow be through a short tube whose section is the same as that of the orifice, it is found that the quantity dis- charged is about 0.82 that of the theoretical, and as the sec- tion of the stream is the same as that of the tube the entire loss is due to a loss of velocity ; hence, for this case V = 0.82 VW^. (263) This reduction of velocity is caused by the interference of the fluid veins within the tube near and about the section where the greatest contraction would take place. If a small hole be made in the pipe, at a distance from the inside of the vessel equal to the radius of the pipe, it will be found that air will rush in, thus showing that there is a negative pressure in the pipe at that point. Different coefficieiits are found for other cases. If the velocity be the same at all parts of an orifice whose section is k, we have for the quantity of flow in a second — Through an orifice in a thin plate : q = 0.64 X 0.97y^ V^ = OMk V2yF: (264) and through a short tube : f;g;'i52. q = O.S2k a/2^ (265) 195. If the orifice is so large as to cause a perceptible ve- locity, Vi, of the free surface, we have where s is the area of tlie contracted section and ^i that of the free surface. This value of Vi substituted in (254) gives: ^ (265a) [1»«.] QUANTITY OF DISCHARGE. 303 If 8 be 80 small compared with ^i that it may be neglected, "we reproduce (261). li s = Si^ v = co, which shows that this condition cannot be realized. The liquid would drop like a free body, and hence its velocity would not be dependent upon liquid pressure. Questions. — 1. If two conical vessels of the same dimensions and filled with the same liquid discharge themselves through equal orifices, one in the base and the other at the apex, will the velocity of discharge be the same when the heads are the same ? 2. In the preceding question, will the vessels empty themselves in the same time ? 196. If the orifice be in the side of the vessel, and of finite dimensions, the heads of the several elementary streams or fillets will be difierent. Let 2 be the head above any point of the ori- fice, then for an element having this head, we el- have Cr V = b ^2^; -1^. where h is the coefficient of velocity ; and for the quantity flowing through this element in one second, q = c \/%j2 . dzdx, where o is the coefficient of discharge ; and for the quantity flowing through the entire orifice, we have A ij Q=cV^[[V^dzdx, (266) integrated between such limits as to include the entire area of the orifice. The free surface is here supposed to remain at a constant height. Examples. 1. In equation (266) let the orifice he a redxingh., h the breadth^ d the deptti^ AB — /<i, AC = h^ ; required the quan- tity which would flow through the orifice m a U7iit of time. 304 EXAMPLES. [1W.I We have Q=cV2gj J Vzdzdx= ^2ghoi^\'^d2, = LV2^J(V-V)» (266«) . o 2. If the upper surface of the rectangular orifice be at the free surface, the opening is called a notch ; required the quan- tity discharged through the notcli. 3.^ Determine the quantity which will flow from a triangu- lar aperture, the apex being in the free surface, h being the base, h the altitude, and the base horizontal. ' Ans. ichhV2gh, 4. In the preceding example, determine the flow if the base be in the free surface. ^^s. ^chhV^^ 5. Determine the quantity of flow, if the orifice be a circle whose radius is r, and whose centre is at a distance h > r be- low the surface. Am. n^ V¥gh [l - 3^ Q'- ^ ©V etc. ]. 6. To determine the time in which a vessel will empty itself of a perfect liquid through an orifice in its hase. Take the origin at the orifice, z vertical, a the area of the orifice, ^the area of the free surface ; then Kdz will be the elementary volume discharged in an element of time, aaodt = ac ^2g2 ,dt=— Kdz, the quantity passing through the orifice in an element of time, and is negative, since t and z are inverse functions, hence 1 ch - Kdz ac^f^g Jo a/* (266J) ri96.) EXAMPLES. 305 It* the section K be variable, its value mnst be found in terms of z before integrating. 7. To find the time in which a prismatic vessel tilled with a perfect fluid will discharge itself through a mere orifice, «, in its base. Ans, ; — . 8. A vessel, formed by the revolution of the semi-cubical panibola, })i^ = ^, about its axis a, which is vertical, is tilled with a liquid to the height h ; to tind the time in which the liquid will be discharged through a small orifice, section a^ .at the vertex. Here fi h ^r- ,. , And the limits are for ^ = 0, « = A, and for ^ = ^, 2 = 0. Ana. t = Ica^g ' 9. Find the time in which a paraboloid of revolution whose altitude is h and parameter j>, full of liquid, will empty itself through a small orifice at its vertex, its axis being vertical. Here ;ry* = njpx. Arts, t = SacV'2g 10. A conical vessel, the radius of whose base is /*, and alti* tude A, is filled with a liquid ; required the time in which the surface of the liquid will descend through half its altitude, the orifice being at the vertex, and the axis verticaL Here and the limits are z = h, and 2 = i/^ ~ '20aoV(/ 306 EXAMPLES. [196.] 11. Find the time in which a liquid contained in a parabo loidal vessel, ^/^ = hz, will descend equal distances h, the flow being through a small orifice whose section is a, at its vertex. . . TthVh Ans. t = — -^:^ The times are equal for equal heights taken anywhere along its axis. This would form a water clock in which equal times would be indicated by equal spaces. 12. A tank whose height FD is 100 feet above the level of the ground is supplied by a 1 inch pipe which communicates with a IJ inch horizontal pipe at the level of the ground, and is fed by a third pipe 2 inches in diameter proceeding from an accumu- lator 3 feet in diameter, with piston, A^ 10 feet from the ground, loaded with 30 tons. Neglecting all resistances, lind velocity with which the Fig. 154. watcr cutors the tank. Equating heads at G and D we have ------^ D A r 1 ( B 30 X 2,240 9 X 3.1416 + 10 + 34 X 62.5 100 +r- + 34; 2g ^from which t? = 64 feet per second 13. Water is discharged from a vertical rectangular oriiice of height d and area A, Show that approxi- where h is the depth of centre of or- ifice below the water level, the orifice being fully immersed- Fig. 155. [197. J DIVERGENT TUBES. 307 From the figure, with the origin at the upper edge of the orifice we have or Developing ( ^^ + q ) ^"^ (^^ ~ o ) *^ ^^^^ terms, and per- forming the operations indicated, we have 14. Find the time in which the liquid in two prismatic ves- sels will come to the same height, one discharging itself into the other, through a short pipe connecting them at their bases. Let h be the area of each base, h their height, A the section of the pipe ; and initially, let one of the vessels be filled, and the other empty. 197. If a conically convergent tube BE oi the form of the vena contracta be attached to the orifice j5, and to the small end ^a tube slightly divergent be attached, it is found by ex- periment that the amount of flow is in- creased, and is even greater than if the discharge be through a simple orifice, except when the flow is into a vacuum. It appears that the liquid adheres to the sides of the Pio, 156. tube, carrying away the particles of air from within the tube, tending to make a partial vacuum at E, or at least to diminish the internal pressure, thereby making more 308 PRESSURE OF [197.^ effectual the head AB and the pressure of the air on the upper surface. The tube being entirely filled, it is the case of steady mo- tion, and Bernoulli's Theorem applies. If jp be the pressure per unit of the atmosphere, jp^ the pressure at E^ at first unknown, and considering the velocity of the surface v^ = 0, equation (253), we have for the head at A (potential) W and for the head at F^ z being zero, and for the head at ^, z also zero, but these heads are all equal, hence A + Z = |l+^ = f+£.; (267) w zg w 2g w if = 2gh; which is the velocity which it would have through a mere ori- fice at E; but as the section at i^is larger than at E, the flow has been increased, and hence the velocity at E has also been increased. This becomes apparent, from the equation ViSi = vS; .:v,=^v, (268) where Vt and Si apply at E, and xS' at 7^ which is larger than ^. [197] FLUIDS IN MOTION. We also have from (267) 2g w or the hydraulic head exceeds A. To lind the pressure j9i, we have from (267) and (268) 309 (269) jp^^jj^wh {\ -'^); and since Sx < S, yve have^i < j). If ^1 = 0, we have sy p -{• wh wh ' (270) (271) and if the liquid be water, p -r- w = ^^ feet, nearly, the head due to the pressure of the air, and the expression becomes (!)■= 34 Since pi cannot be negative for steady motion, equation (270) gives the limiting ratio of the sections, but this limit cannot be quite reached in practice. Eytelwein found that when the mouthpiece BE was shaped like the contracted vein, followed by a divergent tube whose length was 8{| inches long and angle of divergence 5° 9', that 2.5 times as much water was discharged as through a simple orifice of the size of the section at E^ and 1.9 as mnch as through a short tube of the same ^ section as at E. If two vessels be connected by a tube as shown, and filled to the same height with the same liquid, ^»o- ^^ea. and a stream be established in any manner, it will continue to flow across the space when a small portion of the tube is 810 REACTION [198, 190.J removed, provided the velocity be sufficiently great, and will cease only when overcome by friction, or by the difiEerence in heads in the vessels. REACTION OF FLUIDS. 198. Newton's Tliird Law of Motion, being universal in its application, includes the action and reaction of fluids. If a heavy fluid discharges itself through an orifice in the side of a vessel suspended by a cord, the vessel will be forced away from the stream and the cord held in an inclined position. The pressure which would be exerted against the side of the vessel if the orifice be closed, is removed when the orifice is open, and the pressure con- tinuing on the side of the vessel directly opposite the orifice forces the vessel in that direction. FiQ. 156&. There is also a pressure exerted in the same direc- tion due to the deflection of the fluid veins from their course, as will be shown hereafter. The latter is called a reaction. 199. Centrifugal Action. — The force wh^'ch deflects a body from a tangent to a curve is called a ceiitripetal force^ and the equal opposite action of the body upon the curve is called centrifugal force. Strictly speaking, we should say that a force is developed between the body and the curve, which, acting one way against the body, forces it away from the curve, and in the opposite direction produces an equal pressure against the curve. If the body be attached to a cen- tral point by means of a cord, the centrifugal action would be exerted upon the fastenings at the centre ; or, if there be no rigid connection, as in the case of the planets moving about the sun, the centrifugal force would be the same as if the planet moved on the concave surface of a solid coinciding in curvature with the orbit. If the motion be in a circular arc, let m be the mass of the body, V its velocity, r the radius of the path, and cp the cen- trifugal force. Since the centripetal force simply changes the direcdon of motion, if the velocity of the body be constant^ poo.] OF FLUIDS. 311 <p will also be constant. At A the body will be moving in the direction of the tangent AB, and the centripetal force will act in the direction ^ (9 ; so that the body would reach B on account of the motion in the same time that its centripetal force would draw it to D, the points B and D being con- secutive to A, Let and then AOC= 6, AD = X, X = r {1 — cos 6), Differentiating twice, dividing by di\ and multiplying by m gives ^^ j^ = ^z* cos V -T-- . the left member of which will be the value of force deflecting the body from the tangent A B, equation (21). But at ^, 6^ = and dt = GO, the angular velocity. Hence we have (272) where v is the velocitv along the arc of the circle. Equation (272), is the same as equation (141), and also the last term of equation (146). 200. Resolved pressures. — If a particle be shot into a 312 REACTION [200.1 small perfectly smooth tube having a circular bend AEB, it will exert a uniform radial press- ure upon the tube ; the compo- nents of which in a given direc- tion as CX will depend upon the position of the particle. The sum of these components for a given length of arc will be the same as if that length were full of such Fig. 158. particlcs all moving with the same uniform velocity ; and if such a stream of particles be contin- uous, the pressure will be constant. Suppose then that a steady stream of fluid passes through the tube, and let w — the weight of a unit of volume of the fluid, h = the section of the stream, T — the radius of the centre line of the tube, V = the velocity of flow, s = AD = any portion of the path from the initial point of the curve, 6 = ACD, CX parallel to the tangent at the initial point of the curve^ and GY normsil to it. Friction being discarded, the velocity will be unifornk throughout the tube. It is required to find the pressure in the directions CX and CY. For any element of length, we have from the figure, and from (272) Integrating ds = rddy dcp^'-^Ms"^. ' (273) 9 ^ <P = — fCS * ^ g r M—, (2734 where Jfef is the mass flowing into the tube in one second. (aOO.J OF FLUIDS. 313 Resolving dcp, equation (273), parallel to CX and CY re- spectively, and integrating gives Xe = - W f sin eae = ~ hi^ (l - cos 6), (274) ^ Jo ^ Tez^—kv" f cos Ode ='^ki^ sin ^. (275) <7 Jo .^ If the angular deviation be 90°, then 6 = \7r, and (274) and (275) become Xi,=-M, (276) y r».='"-M (277) which are identical. For motion through a semicircle z= n^ and we have X^=2-M, (278) g r. = ; (279) the last of which shows that the pressures normal to the line of the stream balance each other. For the entire circum- ference =± 27Cy hence (280) We observe that equations (274) to (280) are independent of the radius of the path ; hence we infer that the path may have a variable radius, but cannot be zero, since equation (272) will be infinite for r = and v finite. If h be the height due to the velocity Vy then v* = 2gh, and (276), (277), (278), become respectively Xi„ = 2wkh, (281) T^„ = 2wkh ; (282) X, =4wkh. (283) 314r REACTION [201, Hence : Deflecting a continuous stream of frictionless substance through an angle of ^0° along a cwrved path^ produces a press- v/re^ loth in the direction of the initial motion and normal tJiereto, equal to a prism of the matter whose base is the section of the stream^ and whose altitude is twice the Jieight due to the 'velocity. Arid deflecting such a stream 180°, the head due to the pressure (283) in the direction of the initial motion will be FOUK times the head due to the velocity. If Jf be the mass of the liquid flowing through any section of the stream in a unit of time, then — kv = M. 9 (283a) and equations (274) to (278) become Xe = Mv{l - cos 6), (284) Ye = Mv sin 6 ; (285) Xi„ = Mv, (286) Yi. = Mv', (287) X„ = 2Mv. (288) Equations (284) to (288) show that the resultant pressure due to deflecting a fluid from a rectilinear course varies flrst as the momentum of the fluid per second, and second, as a function of the angle through which it is deviated. APPLICATIONS. 201. In the following applications all resistances due to friction, contractions, enlargements, or whorls and eddies in the stream will be discarded. 202. Discharge from the side of a vessel. — In Fig. 1666, considering that the fluid tilaments have their origin in [203, 204. J OF FLUIDS. 315 the free surface, their initial direction will be vertically down- ward, an<l in order to issue Iroin the oritice horizontally- must be deflected throug:h an angle of 90° j hence, equations (281) and (282), the pressure on the oi)posite side of the vessel due to the discharge of the liquid, will equal a prism of the liquid whose base is that of the contracted section, and whose height is twice the head above the orifice. The correctness of this conclusion in regard to the horizon- tal pressure has been proved by one Peter Ewart, an English experimenter, who determined the pressure by direct measure- ment (Memoirs of the Manchester Phil. Soc, Vol. IV.). Fio. 159. 203. Discharge vertically upward. — In this case the change from the initial direction of the motion will be 180° and equations (283) and (279) are ai)plicable, if the section of the vessel A B is sensibly the same as that of the orifice C, in which case the pressure vertically downward will equal the weight of a pri^m of the liquid whose base is the section of the stream, and whose altitude is four times the head due to the velocity'. But when the area of the ori- fice is small compared with that of the vessel, the velocity of the elements in A B will be small com- pared with those in the immediate vicinity of the orifice, and if the former be neglected, the downward reaction will be nearly that due to the deflection through a quadrant of the elements at a sensibly uniform velocity ; and, hence, equal to a prism of the liquid whose base is the contracted section of the stream, and whose altitude is twice the head due to the velocity. 204. If the discharge be from an orifice in the base of the vessel, as in Fig. 100, in which the orifice is so small comjiared with the sec- tion of the vessel that the A'eh)city of the surface may be neglected, then will the reac- tion be due to the deflection of the elements Pig. 160. 516 REACTION [305, 206.| in the immediate vicinity of the orifice, which may be consid- ered as of uniform velocity and nearly all deflected through a quadrant, from a direction nearly horizontal to nearly a vertical direction 5 hence, the vertical reaction will be upward and nearly Y = 2w1chy or the upward reaction will be that due to twice the head producing the velocity. 205. If a stream of liquid impinge normally against a plane surface, in the immediate vicinity of the intersection of the axis of the stream and the plane an eddy or whorl of liquid will be formed, over which the stream will flow as along a curve and be discharged tangentially to the plane with its initial velocity. The direction of motion being changed 90°, Fig. 161."" equations (281) or (286) will determine the pressure exerted by the plane, and we have F = 2wkh = Mv; (289) that is : The pressure exerted hy a liquid stream flowing nor- mally against a fixed plane equals (numerically) the weigJtt of a prism of the liquid whose hose is the section of the stream^ and wJiose altitude equals twice the head due to the velocity^ or equals {numerically) the momentum {per secmid) of the mass imjnng- ing against the surface. The experiments of Michelotti, Weisbach, and others show that this conclusion is very nearly realized when the im])inged surface is at least six times that of the section of the stream, and placed at a distance of not less than twice the diameter of the stream from the oriflce. 206. Cup vane. — If the axis of a stream coincides with that WJ OF FLUIDS. 317 of the axis of revolution of a surface, and impinges against the concave surface, they will be deflected as before, and flow- ing along the eddy as along a curve, the fila- ments will leave the surface tangentially, and equation (284) will be applicable. If the tangents to the surface at C and D are parallel to the axis of the stream AB^ we have d= TT, and (283) or (288) will be applica- PlG. 162. ble, and we have P = ^wlh = 2Mv. (290) The resultant pressuy^e due to the impulse of a liquid stream against a concave hemisphere equals the weight of a prism of water whose hase is the cross section of the stream, a/nd whose height is four times the head due to tlie velocity. Weisbach found by experiments with air impinging against a concave surface that the pressure was about 0.88 of the theo- retical value. 207f Bent pipe. — If the tube through which the liquid flows be bent through an angle 6 at the point B^ an eddy, or whorl, will be formed at the angle, so that practi- cally the flow will be along a curve, and equations (284) and (285) give the resultant pressures parallel and normal to the initial direction of the stream, which being from A toward B will be Fra. 163. X =Mv{l-cose) T--=Mv sin e (291) The resultant 7? will be B = a/X^ -h^V = Mv\/2{1 -cos (9); (292) which alone will prevent the bodily movement of the tube 318 REACTION [207.] were no other external forces acting. The direction of M will bisect the angle ABC. In practice the quantity of liquid discharged through a bent pipe will be less than through a straight one of the same sec- tion, on account of the contraction of the stream at the bend. The mass J/" will be the quantity actually discharged. If two forces, each equal T^ were acting along the branches of the tube and away from the angle, and of sufficient magni- tude to produce the resultant M, we would have 2-3 + 2^2 ^2^2 cose = J^ = 2J!/'V (1 - cos 6), .\T=Mv; (293) which gives the required value of the force T, If there be two bends, jB and (7, in the pipe, let two forces T=Mv, one at A and the other at (7, act away from the "^f\ angle B; they will hold the part ABC. Similarly, one at D and another at B^ each equal to T and acting away from C will hold BCD; but the equal and opposite tensions along BC neutralize each other, leaving the ^-^-^^^SW/ tensions at A and D. Hence, if frictional Fig. 164. resistances be neglected, A perfectly flexible tube of uniform section having bends of amy curvature, and its ends fixed in any position^ will not change its curvature on accov/nt of the pressure due to the flow- ing of a fi/uid through it The effect of the weight of the fluid, which is not included in the above inference, would cause the tube to conform with the plane, or other surface on which it rests. Tlie pressure in a pipe being normal to the carve at all points will be, from Equation (o), page 139, [208, 209.] OP FLUIDS. 819 but from (272) this becomes T= . r = rm?^ = Mv, r as before shown by (293). 208. Impinged surface inclined. — If the plane receiving the iiiipulso be inclined an angle 6 to the axis of the stream, and the stream be confined be- tween guide plates so as to flow along the plane in the line of greatest inclination, the case will be essentially the same as that of a bent tube, and hence the press- ^w. les. ure directly opposed to the stream, and also normal thereto, will be given by equations (284) and (285) respectively ; or F = Mvil-cose)) 209. Kemaek. — Liquids act by impulse when suddenly changed in direction ; and we have seen that the measure of this action is the same as when the stream flows along a curve of finite radius. In practice, however, certain resistances fol- low an impulse, due to various causes, such as the contraction of the stream, eddies, or so-called whorls, which make the effi- ciency of the fluid less than when it acts by simple pressure. Some writers improperly use the term impact in this con- nection, as if the action were the same as that of the impact of inelastic bodies ; but the impact between liquids and solids is only infinitesimal in amount, and hence eludes measurement. The true action is not an impact' but a pressure — an action and a reaction of finite magnitude. The stress between finite solids during impact is rarely sought ; and, indeed cannot generally be found, for the law of action is generally unknown. To find it, the stress as a func- tion of the time must be known, so that the value of fFdt may be found. Also the law of the distribution of the stress 520 EXAMPLES. [209.] throughout the section in contact must be known, from which it appears that the stress may be variable, and the finite vahie sought will be the sum of the infinitesimal stresses acting upon the several elements of the section of contact. Inelastic bodies have a common velocity after impact ; and if a small inelastic body impinge against an indefinitely large one at rest, the motion is destroyed ; but if a liquid impinge against such a body, the direction of motion is simply changed. In the cases above considered, both the momentum and the kinetic energy of the liquid, are the same after the impulse as before. Ex AMP LE8. 1. The nozzle at the end of a flexible pipe of a fire engine is directed at 45° to the horizon, the pipe lying along the ground. What is the apparent increase of weight of nozzle when 150 gallons of water per minute are discharged with a velocity of 80 feet per second ? Also, what is the tension of the pipe ? For the tension we have T = Mv = 4cO lbs. ; the vertical component of which will be Tsin 45° = 28 lbs. ; which is the apparent increase of weight. 2. A jet of water of sectional area A impinges beneath a horizontal plane of weight W. Find the energy of the jet re- quired per second to support the plane? Let m be the mass of a unit of volume of water. Then the momentum acquired per second must equal W, - Av' = m (1) The energy of the jet is wAv X - , (2) 1209.] but from (1)^ EXAMPLKa. ,/w . and substituting in (2), 3. A vessel containing water and weighing 1 ton, within which the pressure is 9 atmospheres, is supported by discharg- ing water downward ; what is the diameter of the jet ? The available head, supposing the discharge against the at- mosphere, is 8 atmospheres = 8 x 34 feet of water. The momentum of the jet, and the consequent reaction m w ^tr'^ 2,240; or ^'x — X 2^ x8x 34 = 2,240; .'. d — 3.474 inches. 4. A jet of water strikes a fixed vane at an angle of 30® and then glances off at an angle of 60° rela- tively to the tangent through the initial point of impulse ; required the resultant pressure on the surface, the jet deliver- ing 600 gallons per minute, with a veloc- ity of 10 feet per second. The entire deflection of the jet will be 90° ; hence R = ViMvf + {Mvf = MvV% Pie. 600 X 8.4 60 X 32i 10 X V'2 = 36.9 Iba. ^22 EXAMPLES. [210.] 5. A jet of water 4 inches in diameter, velocity 25 feet per second, impinges on a fixed cone at its vertex, their axes coin- ciding and the apex angle being 30° ; find the pressure tending to move the cone. .P = — Qv{l- cos I a) = 2— Qv sin^ 7° 30'. = 3. 6 lbs. 6. A jet of water 4 inches wide and 1 inch thick impinges tangentiallj on a concave cylindrical surface of 6 inches ra- dius, flowing over it in a stream of the same section and finally leaving it tangentially after being deflected through an angle of 60° ; the velocity being 10 feet per second, what will be the intensity of the normal pressure, and the resultant force in the direction of the jet ? According to equation {213a) we have for the entire cen- trifugal force where s is the length of the arc of contact, and dividing by the area of the concave surface, which is -^^ s, gives for the iik tensity of the pressure, or 3 X 3Jf-, 62 J 4 1 ... 10 32f ^12^12 ^^^^T"^^*^^'"'''''^'* The resultant pressure in the direction of the jet will be Mv (1 - cos 60°) = 2.7 pounds. 210. If the surface impinged upon be in motion and moves in the initial direction of the stream with a uniform velocity u, the relative velocity will he v — u. If the same mass JbTas before is deflected by the vane, the pressure will be Pll.] REACTION OF FLUIDS. 323 the same as that of a stream moving with the velocity v — ti against a surface at rest ; hence if d be the deflection of the stream relatively to the surface^ we have only to write v — u in equations (284) and (285) to make them applicable to this case, observing that if the surface moves against the stream n will be negative. Hence, we have Xe = M(v- u) (1 - cos 6) = 2M{v - u) sin^ ^6, (295) Ye = M{v - u) sin 6 ; (296) .\X^„=M{v-u), (297) ri„=M{v-n); (298) and similarly for other values of 0, WORK DONE. 211. When the body — called a vane — receiving the impulse of the stream, moves in the direction of the stream or at an acute angle therewith, work is done and energy is imparted to it and the mechanism attaci\ed thereto ; but if the motion be in the opposite direction, energy will be imparted to the fluid. If P be the pressure exerted by the fluid irpon the vane whose velocity is u, the rate at which work will be done — or the Mecha7iical Power* — or simply the Power — will be Pw, (299) where P will be the value given by equation (295) for this case, the entire work according to the supposition, being done in the line of x. Hence (299) and (295) give Pu = Xeu = Mic {V - u) (1 - cos 6). (300) If the deflection be through a right angle, make 6 = 90®; and if two right angles, make 6 — 180°. -^ — i„^ * Author's Elementary Mechanics, Article 99. 324 REACTION OP FLUIDS. [313.1 Equation (300) is a maximum in reference to t« as a varia- ble, when u=iv^ (301) which reduces (300) to Pu = Xe.iv = iM^{l- COS 0), (302) and when the deflection is 90°, this becomes Fu = AV . iv = iM^ ; (303) »nd for 180^ Pu = X„.^v = iMv", (304) Equation (304) gives an amount of work equal to the entire energy of the stream ; if a fluid stream flows into a vane, like Fig. 158, or Fig. 162, where the fluid veins are completely re- versed in direction, and the vane is urged forward with half the velocity of the stream, the mechanical power imparted to the vane will equal the entire energy of the stream, and the efficiency is said to hQ perfect. In this case the fluid leaves the vane with no actual velocity. 212. We will now deduce these results from the principle of the Conservation of Energy. There being no Joss of energy from friction or otherwise, tlie kinetic energy of the stream be- fore entering tlie vane will equal the energy at discharge plus the ^ work imparted to the vane ; or Fig. 167. ^Mv"^ = Fu + iJf F^, (305) where V is the velocity of discharge. Let FGhe the stream having a velocity v, GF the direction of motion of the vane having a velocity u in the same direc- tion as the stream ; then will v — u be the velocity of the stream relatively to the vane at entrance, and since there are [313, 214.] WORK OP IMPULSE. 325 no resistances, it will quit the vane tangentially with the ve- locity BD = v— u, and at the same time it will move for- ward with the velocity BC = u ; hence the actual velocity will be BF= V= V(v- uf +u^ + <2,u{v- u) cos 6 ; (306) which in equation (305) gives Wv" = Pu -\- ^M[{v - uy + u^ + 2w (v - w)co8 6] ; (307) from which we find, Pu = Mu {v — u){l — cos ff), which is the same as equation (300). Dividing by u gives equation (295). From (305) we have Pu = ^Mv^ - i MV\ (308) or, TTie energy imparted to the vane equals the loss of energy of tJieJhiid. 213. Efficiency. — The efficiency of a stream in imparting work to a vane, is the ratio of the energy so imparted to ilie \Lctual energy of the stream. Let e be the efficiency ; then, for the preceding case, we have, equation (308), ' = m^=' — m — ^^ ~ ~^"^ ^^^^^ 214. The Resultant Pressure in Fig. 167 will be P = VXl + r I = M{v -u)2 sin i<9 ; (310) and tan PGP =^' = cot iS = tan (90° - iff) ; .-. PGP = K180° -e) = iFGB. (311) 326 WORK OP IMPULSE. [215] The line of the resultant pressure bisects the angle between the ajpproaching stream FG and the plane AB. 215. General Case. — Let the vane move in any direction, ' Let AB be the vane, CD the initial direction of the stream, having a velocity v = DF^ u = DE— the velocity and direc- tion of motion of the vane ; v^ = BG ~ the velocity of discharge relatively to the vane, which will be tangentially to the vane; Y = BH ~ the actual velocity of discharge, and P the press- ure exerted in the direction of motion DE. There being no frictional re- H sistance, we have, as before, Wv" = Pu + iMV\ (312) Pig. 168. To find Vy let — FDJ= the angle between the stream prolonged and a tangent to the vane at the point of impulse ; q) = EDF, fi = the angle between Z>i^ and the tangent BG, The line J5ZF joining ^and F, represents the magnitude and direction of the stream relatively to the vane at I), and since there is no loss of velocity relatively to the vane while passing along it, we have BG = EF =vi= Vt?' + ^' - 2vu cos cp. (313) Drawing 6^^^ equal and parallel to DE = u^ the line BH '. = V, will represent the actual velocity of discharge, and we /have F^ = v? + u^ - 2v,u cos BGH, = vl-{- u^'-h 2v^u cosX^ + <p) ; (314) which after substituting v^ from equation (313) gives V^ = 1^ — 2u{v cos (p — u) + 2ucos{/3 +(p) Vv^ + u^— 2vu cos - } ; (315) P^^-1 AND PRESSURE. hence, equations (312) and (315) give Pu — Mu[v cos (p — u - COS (/? + (p) ^1? ArU^ - 2VU COS (p] ] (316) m P = M{v COS q) — COS (/? 4- (p) "s/v^ + U^ — 2VU COS cp) if [v COS (p — u — Vi COS (/3 + cpyi (317) If k be the loss of energy compared with the total energy of the stream, the efficiency in imparting work to the vane will be €= 1 Lv Pa iMv'-iMV COS cp — ^■ IT I- ; (318) U cos ifi+<p)^l+^-2}cos.p] from which the condition for maximum efficiency, u being variable, may be found, but the result will be too complex to be of practical value. These equations include not only all the results of the preceding cases, but also several others. They are independent of the initial slope of the vane, rela- tively to the stream, but are dependent upon the relative di- rection /3 with which it quits the vane. 216. Discussion. — Case I. Let v cos (p —u = —v^ cos (^ -f cp). Then equations (317), (316), (315), (318) become, respect- ively, P = ^M {v cos (p — u) Pu = %M(v cos q} — u)u Y^ = -y' — 4 (v cos (p — u)u r • v^l") ^ 4: (v COS q) — n)u 328 WORK OF IMPULSE [216. J In this case vcos (p is the projection of the velocity v on the direction of motion of the vane'w; and in all cases where work is imparted to the vane, the latter will be less than the for- mer, and hence v oos cp — u will be positive, and cos (/? + cp) will be negative, or p+ cp will exceed 90°. The second member, v^ cos (/? + <^), is the projection of v^ on the direc- tion of u\ hence, the assumption makes the resultant pressure (Coincide with the direction of motion of the vane ; also the projection equals EFqo^DEF', hence, cos KEF^ - cos(<^ + y^); .-. KEF = 180^^ - (^ + P). (320) The internal angles at D and N oi the triangle EDN equal the external angle KEN^ or KEN ^cp + P', ,'. DEN = 180° - KEN = 180° -{cp + ^); (321) hence (320), (321), DEN = KEF, as it should, since, as shown above, v^ may be laid off on EF or EN\ and if the angles be all measured from Z^JS' prolonged, we have, equation (321) : The angle hetween the direction of motion of the varne^ and that of the stream relatively to the vane at the point of hnjpulse, inust equal the supplement of the angle hetween the former line and that of the direction of the stream relatively to the vane where it quits it. The line GB produced will not generally pass through Ey but in any case the relation of the angles will be that given above, since a line may be drawn through ^parallel to BG. The speed of maximum efficiency will be found by making the second member of the last of equations (319) a maximum in reference to 'z^ as a variable, which requires that u = \v(tQ)^(p\ (322) 1316-J AND PRESSUKK, which reduces (319) to P = Mv cos (p Pu = i Jfi^ cos* (p V=^ V sin q) e =z cos^ cp (323) and P, Pu, and e will be a maximum for ^ = 0, when F will be zero. Case II. Let the vane he flat and oblique to the stream. For this case /? becomes ^, and by substituting the latter for /? in (315), (316), 317), (318), the required results will be found. The equations will be of the same form as for the general case. Case III. Let the vane he flxit and move no^rmaUy to the vane. For this case ^ = /?, and cp + ft = 90° ; (324) and equations (315), (316), (317), (318) become P = M{y cos q) — u) Pu = M{v cos q) — u)u r^ = v'-2u (t;cos q>-u) T ' (^^^) _ 2{v cos qj — u)u ^ ^ The speed for maximum efficiency will be when u -^ivcoa q}y (326) which reduce (325) to (327) 830 ' EFFICIENCY PlM P = ^Mv COS <p Pu = iMv^ cos^ (p F^ = (1 - i cos^ q)y e = |cos^ (p Case IY. Z<?^ the vane he flat and normal to the stream^ the vane moving at an angle cp with the stream. In this case ^ = 90°, and we have from (317), (316), (315), P = M {v cos ^ — '?^ + sin ^ V^^ + v/^ — %)u cos ^) Pu — M{y cos cp — u -X- %vi\ cp ^v^ + u^— 2vi^cos (p)u Y^ = v^ — 2{v COS (p— u)u — 2u sin cp ^v^ + u^ — 2vu cos (p -. (328) Case Y. Zet the vane he flat and normal to the stream^ and move in the direction of the stream. In this case P = 90°, (p = 0, which in (315) to (318), or (325), (326), (327), give P = M{v-u) Pu= M{v — ii)u F-.-2«.+ 2^^ (329) = U^ + {y — U) • e = 2(- — u)u 1^ I for maximum efficiency, u — iv P = iMv Pu = iMv' ^ \ V' = i^ 6 = ■^ (330) P16.1 OF VANES. 331 Case YI. Let the vatie he a hemisphere moving in the direo tion of the stream,. Then /?=180°, (p=0, and is a sub-case of Case I., hence equations (319) to (323) be- <jome P = 2M{v -u) Pu = 2M{v — u)u V^ = (v- 2uf _ 4 (v — u)u and for maximum efficiency u = ^v P = Mv Pu = iMi/ V=0 e = l Case VII. Let the vane move normally to tJie stream., and /? = 90°. Then (p = 90°, P+ cp= 180% and we have P = M{- u 4- V'i^ + u*) P u = M{- u + Vi^ + u^)u (331) (332) V^ = 'i^ +2 {u - Vi^ -^ u^)u 2( - u + ^/i? + '(J^)u (333) This is, substantially, a sub-case of Ca^se lY., when q> = 90°. 832 HYDRAULIC [217, 218.J HYDRAULIC MOTORS. 217. A hydraulic motor is any device or machine by means of which the energy of water may be utilized. These ma- chines are of various classes, among which we notice the water pressure engine, in which the water passes into and out of the machine through ports, similarly to steam in the steam engine ; water wheels, in which the water flows against floats at the outer circumference of the wheel ; turbines, in which the water passes through the wheel either radially or parallel to the axis of the wheel ; and reaction wheels in which the wheel is actuated by the reaction of the water passing through it. In making an application of the preceding principles, it will be necessary only to describe the general features of the con- struction of each machine, leaving the details to those treatises which make a specialty of this subject. That part of the stream which impinges against the vane of a motor will be called 2, jet to distinguish it from the stream which supplies the reservoir. 218. A SINGLE VANE moviug with a velocity u normally before a jet having a velocity v, will be impinged upon with* the relative velocity v — u, and hence the quantity impinging per second will be rnk {v — u)y and the relative momentum will be F = -k{v-uf; (334) and the work done on the vane will be Fu = —Mv-uyu; (335) which is a maximum for u = \v. (336) It will be observed that this analysis gives the same result [219-221.] MOTORS. 383 «8 substituting —k {v — u) for if in equation (300), and making $ = 90°, The analysis just given assumes that only a portion of the water passing a given fixed section of the jet is utilized in producing work, as will be the cape when only a single vane receives the impulse. A single vane will not be used in practice, but instead thereof a succession of vanes at short in- tervals ; in which case it is assumed that the entire mass of the water in the jet is directly involved in producing work. VERTICAL WHEELS. 219. Vertical wheels are such as revolve in a vertical plane, and hence their axes of rotation are horizontal. The principal classes are Undershot Wheels, Breast Wheeby and Overshot Wheels. 220. The undershot wheel has vanes, floats, or buckets at its circumference for receiv- ing: the water. The water flows under the wheel nearly horizon- tally, and after doing its work upon the vanes escapes through the tail-race. There are two kinds — one with flat vanes, the other with curved vanes. Pio. 170. 221. In undershot wheels with flat vanes it is assumed that the water impinges upon the vanes normally and leaves them tangentially ; which suppositions being very nearly re- alized, are sufliciently accurate in practice, and make equations (329) and (330) directly applicable. Let Q = the number of cubic feet per second of the water discharged by the jet against the vanes ; 334 WATER WHEELS. [231.] Then will 62iQ = the pounds of water discharged per second ; and M = ^^ = 2Q approximately, (337) and the mechanical power imparted to the wheel will be, 2d of Equations (329), m m Pu=^Q{v- u)u ; (338) which may be written : The three terms within tlie parenthesis are expressions of heads due to velocities, and their algebraic sum is called the effective head ; hence ^ The effective head is the head due to the velocity of entrance of the jet LESS the head due to the velocity in the wheel at exit, and also LESS the head lost at the entrance of the jet into the vane, and the tvjo latter must he a minimum that the effective head shall he a maximum. This principle may be generalized and used in all cases; thus : Let II = the total available head of the fall, h = the head due to changes of velocity at the entrance of the float ; hf = the head due to the velocity with which the water quits the wheel ; h" = the head due to frictional resistances ; A'" = the head due to whirls or other causes. Then Pu = m^Q\_H-h-h' ^ h" - h'"] ; (338&) iM.] WATER WHEEIA 335 and " = TO <'^> the horse power developed will be 3i>i X '33,000 ^ ^ ^ ' ^ ^ which will be theoreiically a maximum when the circumfer* ence of the wheel lias one-half the velocity of the jet ; ia which case (339) becomes -^ = 32ix^,200^^* = ^ Qh. (340) where h is the head due to the velocity v. If S be the area in feet of the section of greatest contraction, we have .•.JZP = 0.4558a9/A (341) The theoretical maximum efficiency in this case will be (5th of equations (330)) but this considerably exceeds the value realized in practice. The several losses due to the contraction of the vein, the clear- ances about the wheel through which the water escapes with- out acting upon the wheel, the lack of normal impulse, the imperfect action of a thick vein, and other causes, combine to reduce the theoretical efficiency. Experiments show that a 836 WATEK WHEELS, P29.1 well-made wheel will realize about 60 per cent, of the theoret* ical efficiency, giving for fair practice, equation (339), For a practical niaximum efficiency We have 6i = from 0.30 to 0.38. (343) The power of the wheel is independent of its size ; and hence may be so proportioned as to make a desired number of revo- lutions per minute. 222. The Poncelet Wheel. — M. Poncelet, a celebrated French scientist, improved the undershot wheel by making the vanes so curved that the water upon leaving them would flow backward rela- tively to the vane. For this case, equations (331) and (332) wm^m are applicable, and we have theoretically, and for a theoretical maximum ^^ = IS "^^^^y- (345) If the velocity of the circumference of the wheel be | that of the jet, and the buckets be so curved as to completely' re- verse the direction of motion of the water, the wheel should be perfect, or e = 1, (346) but these conditions not being realized, combined with the [223, 224] WATER WHEELS. 337 losses noticed in the preceding Article, make the probctical ef- ficiency in good wheels about €^ = 0.60. (347) 223. A BREAST WHEEL is ono in which the water is admitted at some point opposite the face of the wheel, and the water retained in the wheel to the lowest point by means of a curved trough, or passage way. The jet will enter the wheel with a velocity, hut if the buckets be pro])erly curved and the velocity of the wheel be properly regulated, there will be, theoretically, no loss from this cause. After the water has entered the wheel it will act by its weight through the remain- ing height. Let h be the head due to the velocity, and ^1 = CD = the height through which the water acts by its weight ; then will the maximum theoretical power be Fig. 172. e>2l^(A + /^i). (348) It' Jl = EFhe the entire fall of the water, the theoretical power will be 62 J (2/7, and hence the llieoretical efficiency will be h + h, (349) Experitnents with the best wheels of this class have given an actual efficiency of €, = .75 to .80. (350) 224. Overshot wheels receive the water at or near their highest part, and retain it in buckets during its descent. If ,338 EXAMPLES. [224.] retained only by buckets, the water spills out before reacliing the bottom of the wheel, and thus produces loss of efficiency ; and if the water be retained by a curved trough, this class of wheels will not dif- fer essentially from the breast wheel in theory. If the water acta by its weight through the height , ,,, . CD - h then will Fig. 173 ^ the mechanical power be Pu =A3I{v - u)u + 62lQh,, (351) where ^ is a coefficient wliose value in the case of flat vanes^ will be unity, as shown by the 2d of Equations (329), and for a complete reversal of direction of the jet will be 2, as shown by the 2d of equations (331) ; hence the real value of A will be more than 1 and less than 2. If the velocity of the circumference, 'w, of the wheel be half that of the water when it enters the wheel, and there be no loss from the impulse, then and we have A (v — n)u will equal AB = h ; Fu= e2i Q{h + h,), (352) the same as (342), and the efficiency of the wheel would be perfect. But there will necessarily be some loss from impulse, and an additional amount from clearance in the curved trough, and still, more from the imperfection of action due to the thickness of the jet, all of which combine to make the practi- cal efficiency of wheels well constructed and properly oper- ated, of e, = 0.75 to 0.80. (353) [224.] EXAMPLES. 339 Examples. 1. Water approaches an oversliot wheel with a velocity of 12 feet per second, the buckets nioviiig with a velocity of 6 feet per second where the fall is 20 feet, hut on account of empty- ing the buckets 4 feet are wasted ; iind the efficiency, and if the supply be 600 cubic feet i^er minute, find the horse power. The relative velocity of 6 feet per second in the race is equiv- (52 9ient to a head h = p— . Tlie heisrlit through which the water 2y iujts by its weight will 1)0 20 — 4 = 16 feet. Then, for useful work per second, we have, making A = 1 'n equation (351), mQ{U - C) X 6 + wQ X 16, 36 — ^+ IGwQ, J The energy at command will be i7» ^ (12)2 + 20m? ^, or 72-^^+20 7/;^; and we have for the efficiency, 30 4-16 X 32t _ ^ 72+ 20 X 32I ' The horse power will be (36 4- 16 X ?,% ) X 62| X 600 _^^^ 33.000 X m\ 2. The section of a stream of water falling vertically over a dam 12 feet high is one square foot at the foot of the fall ; re- quired the horse-power. Anjs, ?>1*%1 + • 340 EXAMPLES. [224.] 3. If a breast wheel whose diameter equals the height of fall, receives a stream of water, section /S', directly opposite its centre, if there be a loss of 5 per cent, of head on account of the clear- ance at the bottom of the wheel, and of 10 per cent, of velocity at the contracted section ; required the horse power and the maximum efficiency. If the vanes be flat, and the impulse normal, the velocity of the circumference being u, equations (329) will be applicable, but in properly constructed vanes, Case Y., equations (331) will be applicable. Assuming the latter, let H be the radius of the wheel, then v = 0.90 V'2gR, and, for maximum efficiency u — ^v, or u = 0.45 V2gli ; and the power of the impulse will be (3d of (332)), if the work done by the weight will be 62i X Q X 0.95JR; hence the total work will be 110^^. (a) The work done by 62J(> falling a height 2B will be 2 X 621QJ2; hence the maximum efficiency will be [224.3 EXA.MPLKS. 341 To find the horse power, substitute m equation ((/), and find 110 X 7.2 VA' . .^ X ^ X 60 {c) HP 33,000 1.446^^'. 4. Required tlie horse power and efficiency of an under- shot wheel when the velocity of its circumference is \ that of the jet ; the quantity of water discharged being 600 cubic feet per minute. 5. Required the horse power and efficiency of an undershot wheel when the velocity of its vanes is | that of the jet, when Q = 600 cubic feet per minute. 6. If the discharge of water against a Poncelet wheel be 100 cubic feet per second, and the velocity of its perimeter be | that of the jet, required the horse-power and efficiency. Y. Determine the horse power and efficiency in the preced- ing example if the velocity of its perimeter be J that of the jet. For the horse power we have, equation (344), z= r^iT^ nearly. For the efficiency we havb ^~ w^ ~^* 8. If the velocity of the perimeter of a Poncelet wheel be 5 342 EFFSCTIYE HEAD. [225.J feet per second, and the discharge of the jet be 200 cubic feet per minute; determine the section of the jet for maximum efficiency. 9. In the preceding example, determine the section of the jet when the efficiency is one fourth the theoretical. 10. In the Poncelet wheel, the velocity, u, of the circumfer- ence being fixed, and the required horse power, IIP, being given ; find the head due to the velocity when the efficiency is a maximum, and the section of the orifice, the coefficient of contraction of the vein being 0.62. 11. At what velocities of the wheel wiU the efficiency be zero ? 12. If a wheel is being run at a velocity to produce a maxi- mum efficiency with a jet of 25 square inches, how much must the section of the jet be increased to produce the same work when the wheel revolves with | the former velocity ? 225. It will be seen from the preceding discussion that if the water can be made to enter the wheel or act upon its vanes, without shock and leave it without velocity, the highest ef- ficiency will be produced ; and if in addition to these there be no losses from friction or clearance, or other imperfect condi- tions, the efficiency will be unity, and the wheel is said to be perfect. Some writers distinguish between impulse and reaction ; the former being applied to the direct action of the water in producing the impulse; and the latter to the effect due to the change of motion after the water has entered the vane, and hence may be affected by friction and the direction with which it leaves the vane. According to the view presented above, the action of liquids, whether of impulse or pressure, is of the nature of a reaction. 27ie effective head is that portion of the actual head which would produce the work done by acting upon the wheel with- out loss of any kind. Thus the coefficient of M in equations (300), (316), (323), (325), etc., divided by g, and {v - u) u t326.] BEACnON MOTORS. 34^ -5- ^ in equations (338) and (344) are eflfective heads for the respective cases here cited. EEACTION MOTORS. 226. Jet Propeller. — If water issues horizontally from the side of a vessel, it will, by virtue of its reaction, move the vessel in the direction opposite to that of the jet, when the resulting pressure is suf- ficient to overcome the resistances. The vessel may move with a uniform veloc- ^j^ u w "\J P ity, or with an acceleration eitiier posi- ^^MW^kdw/M tive or negative d impending upon the ac- *"'«• ^^^• tion of the external forces. In this analysis assume that the quantity of water in the vessel is sensihl}^ constant during the action, and discard frictional resistances of the issuing jet. Let h = AB = the static head above the orilice B, V = the velocity of discharge due to the head A, u — the velocity of the vessel, f — the acceleration of the vessel, Ai = a head which, statically would produce the same pressure at the orifice B as is produced by the acceleration, Ui = the velocity produced by the head Aj, V = the velocity of discharge relative to the orifice ; then, ,/ = 2gh, ni- = 2fhi, (354) and F2 = v^ ± V. (355) The actual velocity of the jet will be and the rate of doing work at the instant the velocity is u will be Pu = Jf(V'y' ± V - u)u. (356) 344 REACTION MOTORS. [227.J If the velocity of the vessel is uniform, or the acceleration is so small that it may be neglected, then u^ = 0, and Pu = M{v- u)u, (357) which will be a maximum for u — ^v, in which case the maximum power developed will be Fu = \Mv^: (358) The efficiency will be which has no maximum. But if the vessel on the arm be connected by a pipe to a supply chamber about the axis of rotation, the pressure at tho orifice will be increased on account of the centrifugal force of the water in the pipe due to the rotation. To find this pressure, and the equivalent head, assume that the inner end of the pipe is at the axis of rotation, BC, Fig. 175, and let a — GF^ the distance of the orifice from the axis, m the mass of a prism of the water whose base is unity and length unity, and p a variable radius, then will the mass of an element of length be m dp and the centrifugal force will be ea \ mdp • Q^p — \mQi?f?» Jo If Aj be the head which would produce this pressure, thea /. GD^p" := %j1h = u\ (360) since oop •=. u ; ... F2 = ^2 + u\ (361) Tu = Jl/"(V^ + u^ - uy. (362) 227. Barker's (or Seguin's) Mill consists of a vertical hollow shaft communicating with hollow transverse arms* [227.] REACTION MOTORS. 845 The water is admitted into the upper end of the hollow shaft, and passing downward, flows horizontally into the hollow arms, escaping horizontally through orilices near their extrem- ities, one orilice being on one side of one arm, and the other on the opposite side of the other arm. The deflection of the water from the vertical to the horizon- tal direction will cause both a hori- zontal and vertical pressure, as shown in Article 200 ; but the horizontal pressures will neutralize each other, while the vertical pressure will be resisted by tlie support, and the water will flow into the arms with a velocity unaffected by these pressures, and, if ^ the arms are stationary, the water will "^^^^^^^^fc^^ be discharged at the orifices with a fig. ns. velocity due to the head in the vertical shaft. l>ut when the arms are rotating, the water in them will be forced radially outward on account of the centrifugal force developed, and thus the head of discharge relatively to the orifice will be increased. As the water approaches the orifices which are near the ends of the arms, it will be deflected through an angle of 90°, thus producing a pressure radially outward, and also transversely to the arms, the former of which will be resisted by the solid parts of the machine, and the latter will produce the rotary motion. The two reactions — one at each end of the outer extremities of the arms — produce a couple, the mo- ment of which will equal the moment of the resistances over- come by the machine. Let V = the velocity due to the head h = OB, a = CF, GO = the angular velocity of the arms, u = the actual velocity of the orifices O and /^ = V = the velocity of discharge normal to the arm relik tive to the orifice, Ai = the head due to the velocity u. 846 REACTION MOTORS. IW7.J Then, as shown in the preceding Article, F^ = u^ ^ v^ = d^od" + 7?\ (363) and the actual velocity of discharge will be Y-u=^Ja^a? ^'i^^aGo\ (364) dnd the pressure due to the reaction will be and the mechanical power resulting, will be Pu = MiVa^'oo^ -\-v^- aoo) aco ; (366) and the theoretical efficiency will be M{^/d^(ju^ + t^ -- aGD)aGo 2a ft? V + aoj^ (367) which has no maximum in reference to a? as a variable, but approaches unity as a limit as &? is increased indefinitely. In practice it is found that the efficiency is a maximum when aoo is about equal to v ; and making aao — v, and neglecting all losses, except that due to a loss of actual velocity of dis- charge, we find for the efficiency, 6 = 0.82+. (368) If there were several orifices at distances a\ oJ\ a"\ -etc., from the axis of the shaft, producing reactions P'^ P\ P'", etc., giving expressions similar to those in equation (36£>), and R the resistance overcome with a uniform velocity w at a dis- tance J from the axis of the same shaft, then w^ould the mo- C227.] REACTION MOTORS. 347 ments of the reactions equal the moment of the resistance overcome, and hence Rh = Fa + P"a" +, etc., = M' (a/<^'^o^ + 1^ -^ a'co) a' + M" {y/a'^'o^ ^^- a"(M>) «"+, etc. ; (369) and the mechanical power will bo EhoD = M' (V^'^G^ + i;2 - a' GO) a' GO +, etc. (370) An analysis of the second member of the last equation gives the following : — the quantity a/<^^^ + v^ — a'oo is the actual velocity of discharge as shown by equation (304) ; M' {^/a^(^ + i;^ — a'oD) is the momentum of the mass M\ as shown by the definition in Article 27; and WiVa'^oJ^ + v* — a'oD) a' is the moment of the moirientum, as shown in Ar- ticle 160, or in Article 168 of the Elementary Mechanics, Hence the entire expression, M'{\^a"^iJ^ i- i^ — a oo) a' cd, is the moment of the momentum of the mass M' multiplied by the angular velocity^ and similarly for the other terms. Separat- ing the expression into other terms, we have 3T{Vct^(^ + ^)0''y which is the moment of the momentum of the mass M' having the velocity '^a'^oi? + i;^ in reference to the orifice ; and M'a'oD'a\ which is the moment of the momentum of the mass M' having an actual velocity a'oo, equal to tire actual velocity of the orifice, which, being in a direction opposite to that of the discharge, will be negative. Each of these nnil- tiplied by the angular velocity, a?, will give the corresponding mechanical power developed ; and similarly for the other terms containing M'\ M"\ etc. In this case the water enters the arm without angular velocity, still the principle shown in Article 160 is general, for the effect produced is the result of the mutual action and reaction between the water and arm. Hence, 348 EEACTION MOTORS. [228.] The worh done hy the reaction of water in passing through a motor equals the difference of its moment of moment/am on, entering and quitiMig the wheel. This machine has been the subject of many improvements. To remove as much as possible the resistance of the v^^ater in passing through the arms, sudden turns, eddying, etc., the arms have been made large where they join the body of the shaft, and gradually contracted and curved as they pass outward toward the openings. Instead of arms, there may be a disc having openings which permit the water to escape tangentially, which form is known as Whitelaw's Turbine. In the form shown in Fig. 175, the pivot at the lower end of the shaft supports tlie mill, and all the water running through it, thus producing Fig. 176. much friction. This objection has been in part removed by introducing the water into the mill from the under side, thus producing an upward pressure, and counter- balancing, in part, the weight of the machine. But notwith- standing all these improvements, the mill has been superseded by other more efficient motors. 228. To determvne the pressure due to the centrifugal force. — Let A be the uniform cross-section of the arms, gj the angular velocity, p the pressure per unit, m the mass of a unit of volume, p any distance from the axis of rotation, the initial and terminal values of which are r^ and r^ respectively^ then will the mass of a transverse section be mAdp, and the centrifugal force will be 7nAdp-o^pf "which will equal Adp\ Adp = mAoc^ pdp; [238.] REACTIOX >[()TORS. 349 or 2> — po = \ni(jD^{^i\ — 7'?), (371) which is the pressure required. If w be the weight of a unit of volume, we have ^ iw - w) = 2y C^ - -^o) = <^(^i- ^)^ (372) where h is the head due to the velocity rgCe?, and ^o that due. to t'lGj. Tlie velocity due to the difference of these heads M'ill be u = V2(/ {h - Ao) = goVtI^^, (373) The pressure per unit, j9, will be independent of the form of the longitudinal section of the arm, provided it be con- stantly full. Assume the axis of the arm to be in a hori- zontal plane, and the transverse sections variable, then when stationary Bernoulli's Theorem, equation (253), would give the pressure p at any distance r from the axis of rotation ; then if rotation exist, the pressure at that point will be in- creased an amount produced by the centrifugal force between the distances r^ and r. Letting j?i, Vi, r^, be one set of con- temporaneous values, a.ndp,v,r, another set ; finding — from equation (253) after making z = %, and adding thereto the head due to the increased head given by (372), we have for any point of the rotating arm, ^ -^ 4. !!l _ !^ 4. ^ _ ^ r374^ w" w ^ 2g %j "Ig 2g ' ^ ^ 1. Find the horse power, number of revolutions per min- ute, gallons of water discharged, and efficiency of a Barker's mill, neglecting friction. Area of nozzles 1 sq. ft., radius 3 feet, total head 30 feet, speed of orifices |ths that due to the head. u — \^J%j X 30 = 32.96 feet per second; 27r7i X 3 = 32.96 x 60 .-. 71 = 104.6; 350 TURBINES. F = V^? + 2^, or F= 54.8 ; .Efficiency — U— = 0.75. Horse power = 140.7. 2. A rotating wheel receives 2 tons of water per minute at a radius of 2 feet, and delivers it at a radius of 3^ feet ; on entrance, tlie water is rotating with a velocity of 14 feet per second, and on delivery it is rotating in the opposite direction with a velocity of 4 feet per second. Find the couple exerted on the wheel, and if the wlieel makes 200 revolutions per minute, lind the ZTP developed. 2 X 2240 Moment of morrientuiii on entry = 7-^ -^- x 14 x 2. OU X oL 2 X 2240 Moment of momentum on delivery = x 4 x 3 J. Sum = Couple = 98. rro 98 X 2;r x 200 „ - ^^= 60x650 =^-^' TURBINES. 229. The general definition of a turbine is a water motor rotating about a vertical axis ; but it is usually restricted to those horizontal wheels in which the water is conducted to the vanes by curved guide-plates. They may be divided into four classes : 1. Outward flow turbines, in which the water flows hori- zontally outward from the central shaft, and is discharged at the outer circumference of the wheel. 2. Inward flow turbines, in which the water is admitted at [230. J TURBINES. 861 the outer circumference and flows horizontally inward toward the central shaft. 3. Parallel flcm turbines, in which the water flows down- ward (or upward) through the wheel. 4. Mixed flmjo turbines^ in which the flow may be com- pounded of the Ist or 2d with the 3d above. 230. Fourneyron's Turbine, invented about the year 1827, is a good type of an outward flow turbine. The central shaft Fig. 177. 6^ of the wheel, to which the driving mechanism is attached, passes down through the supply chamber GU^ from which it is separated by the tube LL. 352 TURBINES. [230.J The water passes downward between- the guide plates oo^ and outward between the same, through the gates MM^ thence against and along the vanes AB, finally escaping at the outer circumference of the wheel. The vanes curve backward so as to cause the water to be discharged in a direction opposite to that of the wheel, relatively to the vanes. Tlie vanes are attached to the shaft C by means of the rigid nvm^ff. To find the mechanical power expended upon the wheel, let, in Fig. 177, V r= tlie velocity of the water as it enters the vanes, y = FQR = the angle between tlie tangent PQ io the guide plate, and QR, the tangent to the inner rim of the wheel, Vt = the tangential component of the velocity, Vr= the radial component of the velocity, ^1 = OQ = the internal radius of the wheel, a = angle between the direction of the water as it escapes from the wheel, and XxS' the tangent to the outer rim of the wheel, v', v/, v/, 7*2 = I'espectively, the velocity of discliarge, its tan- gential and radial components, and the radius of the outer rim. Then, Vt =^ V cos y, Vr = V sin y ; {^^^) vl = f'cos a , Vf.'= f'sin a . (376) The moment of the momentum of the radial velocities will be zero ; hence, the moment of the momentum of the water as it enters the wheel will be Mnvt , and as it quits it, it will be Mr^Vt ; hence, Article 227, the work imparted to the wheel, neglect- ing friction, contractions, eddying, etc., will be Fu = M{r,v, - r^vt') w ; (377) P30.] TURBINES. 363 and if h be the head in the supply chamber above the point of -discharge, the efficiency will be ^^{r^U^r,v;)oo ^ (378) Equation (377) is a fundamental one in the theory of tur- bines. The velocities v and v will generally be independent of each other, depending upon the form of the guide plates, the vanes, and the speed of the wheel. For the highest efficiency, the water must quit the wheel with no velocity. There will, however, practically, always be a radial velocity, but when this is reduced to a minimum by the proper construction of the vanes, a maxiinum efficiency will be secured by running the wheel at such a speed as ta make the velocity of discharge a minimum. Assuming it a minimum when the discharge is radial (approximately true), we have v^ = 0, (379) e = r^. (380) The coefficients of M, equations (377) and (379), divided by ff, are the effective heads for the respective cases. If to the effective head there be added the head due to the loss of velocity, there being no frictional resistances, the sum will be the head in the supply chamber. The only head lost in the case of maximum efficiency will be that due to the radial velocity of discharge. For this case, or, equation (376), will be 90°, and v/ = v'. But if w be the velocity of the water along and relatively to the vane, and w^ and Wr the relative tangen- tial and radial velocities, respectively, ft = the angle SXT^ Fig. 177, between the tangents respectively to the outer rim of the wheel and the vane ; then ^'V (381) Wr = Wt tan fi ; \ 28 554 TURBINES. £380.1 the head corresponding to which will be ^? tan^ (i-^2g\ hence, for the case of maximum efficiency, h = (2rtViOo + wj tan2 ^)-i-2g; (382) from which we find that the velocity with which the water enters the wheel is V = Vt sec r={-^^—^I,r:^)^<^r- (383) Hence the velocity of entering the wheel varies inversely as the velocity of the inner rim. As limiting cases assume that y — 0, and /i = 0, and we; have v=-^. (384:) li TiGo = V, then v'=gh, (385) or the velocity will be that due to one-half the entire head. If Tioj = ^v, then i/ = 2gh, (386) or the velocity will be that due to the head. v" = 4:gh, or the velocity will be that due to twice the head, which is an- hypothetical condition. The above analysis is equally applicable to the other classes of turbines ; but the construction must be different for each class in order to realize the conditions here imposed. Thus it will be found that /3 must be larger for a parallel-flow turbine than for an outward flow, and still larger for an inward flow, if the transverse sections of the passages through the wheel are uniform. Its value may also depend upon Ti and r^ a^ will be shown. [231.] TUEBINES. 355 231. Special Cases. — 1. In the Fourneyron turbine the initial elements of the guide plates are radial, and passing out- ward from the axis of the supply chamber, are gradually curved until they finally terminate as nearly tangent to the inner rim of the wheel as practicable, allowing sufficient space between the plates for the passage of the required quantity of water. The least value of y, tan K = — , (38V) gives the greatest possible tangential velocity to the water when it enters the wheel. The initial elements of the vanes are also radial, but in passing outward they are curved backward, and at their terminus make a small angle, /?, with the outer rim. These conditions give w. tan/? = ^. (388) to, In order that the water shall enter without impulse, the velocity of the inner rim of the wheel must equal the tangen- tial velocity of the water, or, Vt = 7\oD ; (;5;-s9) in which case the initial velocity of the water relatively to the wheel will be the radial velocity v^. It will then pas3 along the curved vanes and be finally discharged backward relatively to the vane with a tangential velocity v;<, but the vane has a forward motion of r^oo = nrioo ; hence the actual tangential velocity of the water will be vl = Wt — nriGo, (390) which for the best result must be zero, for which condition we find, combined with equations (388), (389), and (387), tan /i tan ;/ ' .-. tan i/ = 7i^^ tan/?; (392) 356 TURBINES. [231.) which establishes the relation between the outer angles of the blades and vanes relatively to the respective rims of the wheel. The radial velocity through the wheel may be made constant when the wheel passages are full, centrifugal force being neglected, by making the transverse sections of these passages uniform ; and since their breadths vary as the radii rj and r^y the depths must vary inversely as the same numbers, thus making the axial sections hyperbolas. This being done, v^ = iOry and we have tan y = n tan /?, (393) which is the condition commonly assumed. These conditions being realized, we have for the Fourneyron type — frictional losses being abstracted — equations (380), (382), (389), and (390), - = 2+An^/? > ^'''^ ,^ 2gh (395) ^* 2 + ?v' tan2 ^ and from (379), from which it appears that the efficiency is greatest when /? ia least. li J3 =0; then e= 1 and v = Vgh ; and the head due to the velocity with which the water enters the wheel is one-half that due to the head in the supply chamber. Since the water enters the wheel with a small velocity relatively to the vanes, Vt tan y, and is discharged with a greater relative velocity, w = nrico sec /3, it follows that from the point of entrance the velocity of the water relatively to the vmie is accelerated to the point of discharge, while the actual velocity diminishes along the same path. 2. In the cup vane turbine the guide plates are essentially the same as the Fourneyron type ; but the water issues from them [231. J TURBINES. 367 with a velocity due to the head in the supply chamber, and hence enters the wheel as a jet. The vanes are vertical, and their surfaces are as nearly semi-cylindrical as the circum- stances of construction will permit. To avoid loss from shock, the initial elements of the vanes should be tangential to the direction of the water relatively to the vanes as it enters them. Let y' be the angle between the initial element of the vanes and the inner rim of the wheel, v' the velocity relative to the vane at entrance and which will be made uniform by making the normal sections of the stream uniform, and V the actual velocity of discharge, then V cos y ^ v' cos y' = r^oo^ v' sin y' = V sin y, nriGo — V cos ^ = tang. vel. at exit, v' sin /3 = radial vel. at exit, Y =z (nrioo — v' cos pf + v"^ sin^ yff , which should be a minimum for maximum efficiency. As- sume either y or y as known, v being known, and find V a function of <w, and the value of oo that will make Y a maxi- mum will make known y or y\ and v'. Instead of reducing this case, we will take a special one, assuming that the veloc- ity of the inner rim of the wheel is equal to one-half the tangential velocity of the water, and that the discharge from the wheel is radial ; then r^Qo — ^ cos Y^ (397) Vr — V sin y^ (398) tan y' = JL^HUl = 2 tan y, (399) « = i^V cos2 ;/ + 4 sin2 y, (400) nr^oo = Ti cos /5 ; (4:01) for which case P cannot be assumed, but may be found from the last equation. 358 TURBINES. [281.] The energy imparted to the wheel will be, equations (379), (397), Fu = MvirVx GJ = Mv cos y-^v cos y, = W^ cos^ y. (402) The kinetic enei'gy of the water as it leaves the gates will be, W being the weight of water flowing per second, Wh=iMv^', (403) hence the theoretical efficiency will be ^Mi^ cos^ 1/ e = iMv' = cos^ y. (404) 3. In the inward jlow turbine the initial elements of the guide curve are at the outer circumference of the supply chamber, and curve so as to form a small angle with tlie outer rim of the wheel. The outer elements of the vanes are radial, and as they approach the inner rim of the wheel, they curve in a direction opposite to that of the guide plates. Since n^ in equation (393), will be fractional, the angle ^ will exceed y. Tiie water should enter the wheel with a tangential velocity equal to the velocity of the outer rim. The remainder of the analysis is the same as for the Fourneyron turbine. If the vanes are cup-shaped, the analysis will be the same as the second case above. 4. In the parallel flow turbine^ the initial elements of the guide plates are vertical, and are curved as they pass downward so as to terminate as nearly horizontal as possible. The initial elements of the vanes are usually vertical, and the vanes curve in an opposite direc- tion from that of the guide plates, termi- nating at the same angle with the horizontal. This last is a result of making ?i = 1 in equation (393), giving Fig. 178. 1332.] TUBBINES. 359 5. If, in an outward fiow turbine^ the vanes make an angle with the inner rim, exceeding 90° and less than 180°, which will be an angle greater than for the Fourneyrou type, and less than for the cup vane, we shall iiud rco> ^/gh and < V^gh, for a speed of maximum efficiency. 6 An unlimited number of turbines having different forms of vanes may be devised which will give, theoretically, higli efficiencies ; but if friction, whirls, contractions, etc., be in- volved, those forms will be, practically, most efficient, which for the same theoretical efficiency, have the least of the above named resistances. 232. General case. — In any hydraulic motor, let H be the total available head, h the loss of head due to location and form of wheel, or that portion of the head below the wheel (and possibly above) not utilized, Ii the loss by shock, h" the head lost on account of the velocity of the water at quitting the wheel, A'" the head lost from contractions, whirls, friction, etc., then will the mechanical power be Q^Q[H-h- h' - h" - A"'] ; (405) and the gross efficiency of the motor will be H-h-K -h" -Ji" H (406) The efficiency of the motor as a machine may exceed this, for it may be so constructed and placed that its entire efficiency will be developed with a less head than H\ as, for instance, in the case of an overshot wheel, if there were 4 or 5 feet fall below the lowest point of the wheel, it would in no way affect the efficiency of the wheel, but the wheel would fail to utilize the power of the water. In determining the quality of the wheel as a motor, only ^o much of the head should bo 8i^ TURBINES. [282.] considered as is actually necessary for operating the wheel in the place where it is established. EXAMPLES. 1. A stream delivers 10,000 gallons of water per minute to a Fourneyron turbine. If the fall be 100 ft., tan /? = J, ?i = 1.4, and revolutions per minute = 240 ; find the internal and external radii for the most efficient speed, the HP^ and the depth of the casing at the outer periphery. From equation (395) we find o , /2 X 32| X 100 ^^ = ^^^^-f -2 +(1.4)^xi > or ri = 2.14 feet. r^ = n7\ = 2.996 feet. From (394), = .90. innno v 9? EP = 10,000 X 231 X 62i x 100 x .9 1,728 X 33,000 = 227.8. From (391), Wr =^ Vr = nr^GD tan /?, or Vr = 25.1328 feet per second, and 10,000 X 2 31 X 12 _ ~ 1,-728 X 25X33 x 6;r x 60 = .56 inches. 2. An outward flow turbine is supplied with water having a head of 350 feet. Tlie internal diameter is 15 inches, and the external diameter is 21 inches. Angle of guiding plates, [232.) TURBINES. 861 ;^ = 16°, and the vanes of the wheel are placed at the corre- sponding angle. Depth of wheel at the inner periphery, 2 inches. Find the number of revolutions per minute required for the best efficiency, neglecting frictional resistances. Taking account of frictional resistances, and supposing the true efficiency to be Jths, find the best number of revolutions and UP developed. From equation (395), , /2 X 32i X 350 .^, . , , '^'^y 2:0822 "" ^ ^^^ second. 104 X 60 . .on 1 ..• • * -= — = 1,589 revolutions per mmute. Ati X -J If we take frictional resistances into account, and suppose the wheel to be running at its greatest efficiency, we will have for the total energy, mgQh = mQtr} + '-^ v'.n^ tan« /? + ^ 2Fvj (1 + n^ tan' /3). Therefore the efficiency is or 2 + ii" tan^ fi + :2F{^ + n^ tan« fi) ' v[ = J^A, or Vt = 91.9 feet per second, ^1= ~ — = 1,440 revolutions per minute, J;r X -| Wr = Vr — Vi tan y, Wr = 26.35 feet per second. Area of flow at inner periphery = .6545 square feet; .6545 X 26.85 x 350 x 62.5 x 60 x .75 HP 33,000 = 514.45. 362 ' RESISTANCES. [2;">3.] 3. Which requires the larger gate opening for the same work, the Fonrneyron turbine of Case 1, or the cup vane tur- bine of Case 2 ? 4. If the head in the supply chamber be 15 feet, find the velocity of the inner rim for best efficiency in Oases 1 and 2, when /3 = 0, and when fi = 20°. 5. Explain why the same quantity of water flowing through the cup vane as through the Fonrneyron turbine produces the same work, when the velocities of the water on entering the wheel and also the velocities of the wheel are different. RESISTANCES. 233. Sudden enlargements. — If a stream flows along a pipe liaving a sudden enlargement, it suffers a loss of velocity. C If y be the velocity in the small pipe, and Vz that in the enlarged part, it might seem that the head lost would be (v^ — vl) -^ 2g ; but this is true only when the internal forces are functions of the distance between * the centres of force, Article (151), a condi- tion not realized in this case. The particles act upon each other according to a law not definitely known. It may be partly of the nature of friction, but chiefly it results in the production of whirls at the angles of the tube. It is also known that the reduction of the kinetic energy develops heat, and this, being retained in the tube, partially maintains the head, so that the loss is not as great as it otherwise would be. The action is not exactly of the nature of the impact of flnite bodies, but resembles it in regard to the non-elasticity of the water and the comparatively sudden reduction of the velocity ; and for these reasons we analyze the case as for inelastic im- pact. In all cases, however, when the law of action is unknown, the result should be checked by direct experiment. We may assume that the mass m passing out of the small tube in a unit of time impinges upon the mass M in the enlarged {284. J RESIST.A^'CES. 86S part of the tube ; then, according to equation (42), the loss of energy will be im(v-v^)\ (407) and it is found that this expression represents with sufficient accuracy the results of experiment. This amount of energy transformed into heat would raise a pound of water a number of degrees in temperature given by the expression where 772 is Joule's mechanical equivalent of heat. The head lost will be ^- (*««) 234. Resistances m long pipes. — The principles of work and energy on which Bernoulli's theorem is founded may be extended to tlie case in which there is resistance along the stream between A and B, as is the case in actual tubes. The law of the resistance can be determined only by experiment. It has been found that, with sufficient accuracy, the resistance varies as the perimeter of contact between the liquid and pipe in a transverse section, called the wetted perimeter, also as the square of the velocity, and a factor/ dependent upon the con- dition of the pipe. Hence, if B be the resistance at any sec- tion, 8, w the weight of a unit of volume of the fluid, and g the wetted perimeter, we have R = fwcr-^ . 2^ If the velocity and sections be uniform, then will the loss of head for a length I be 364 RESISTANCES. [^5.] The value of/* for rivers, as given by Eytelwein, is where /=« + -, a = 0.007164, h = 0.007409. (411) 235. General case. — Let AB be a stream flowing from a reservoir whose upper surface is at C\ and whose exit is at B. Or, to generalize it, let B be any point in the pipe. Through B draw a horizontal GB, then will CG = BD be the total c. d H 1 m F A k M N L B Fig. 180. head on J5, neglecting the pressure of the air as we may do in most practical cases. If B be the exit, let BL be the head necessary to overcome the pressure at the orifice. If the opening be the full size of the pipe, BL will be zero ; but if it be contracted, it will be of finite value. Lay oft' on BDy the following heads : DII = Ai = — = the head due to the velocity of exit, if HI = \ = the head lost due to the resistance of entering the pipe at JL, IM = /i^ = the head necessary to overcome the resistances along the straight portions of the pipe, MN = hi — head lost by bends in the pipe, NO = hi = head lost hj sharp angles. 1235. J KESISTANCES. 365 OL = Ji^ = head lost by enlargements and contractions, LB = A7 = head due to pressure at B, DB =^ H= total head at B. Then //= h^ + ^2 4. ^3 + ^^ + ^5 + h^ 4. h^, (412) The values of these several heads, with the exception of the ■first, are determined by experiment. They are usually de- termined as multiples of Aj = — , and hence may be written with the use of corresponding subscripts, thus H= (1 +/, +/, +/, +/, +/. +f,)'^. (413) The following are some of the principal values of the coeflScients as determined by experiment : Orifice in a thin plate, which would be applicable when the length of the pipe is zero, f, = 0.054. Straight, short cylindrical mouth-piece perpendicular to the side of the vessel, /a = 0.505. For the resistance along the pipe whose length is ly section «, wetted perimeter c, Wherey is dependent upon the dimensions of the pipe, and according to Darcy, its value is /=K'-^> 366 RESISTANCES. [235.1 Where d is the internal diameter of the pipe in inches, and Ic for clean cast-iron pipe, is Ic = 0.005 ; and for pipe very rough from sediment, k^ 0.01. According to Weisbach, for clean cast-iron pipe, 0.0043 / = 0.0036 -h ^/v where v is the mean velocity. The larger computed value of f would ordinarily be used. For a bend through an angle % the radius of the cni've being r, diameter of the pipe d. A = Fig. 181. i] 0.131 -.1.847(1)*}. For a knee, or sharp angle, % /g = 0.946 sin^ | + 2.05 sin* |- , and if the knee is 90°, this becomes /s = 1 nearly. The value of f^ may be computed approximately, and f^j is dependent upon the condition of the opening at exit. These values in (413) enable one to determine v\ ov v and the other quantities being known, H, the necessary head, may be determined. Indeed any one of the many quantities may be found in terms of the others. To find the pressure on the pipe at any point, as h, lay down from CD the head dh due to the velocity at J, hi the head lost at entrance A^ im the head lost by the resistance of the straight part of the pipe from A to J, mn the head lost by 1235.] RESISTANCES. 367 bending between -4 and 5 due to bends, no that lost by angles, ol all other heads lost ; then will Jh =-jpre88ure onjyvpe at h. The head hg is potential in reference to £G, The line JTL limits the upper ends of the heads due to pressure on the pipe ; and if the pipe lay along this line there would be no pressure except on the under side due to the weight of the fluid. If an orifice be made at b, the liquid would flow out in a stream rising to the height hi, less the resistance at exit from said orifice and the resistance of the air. But if the pipe lay along i^Z, the liquid would not flow out of an orilice at L E X A M p r. E s . 1. Let it be required to deliver water with a given hydraulic head ^q? such, for instance, as is necessary to drive an engine. Then 2(/ 2(/ -^ d'lg' where '2F\^ the sum of the coefficients due to valves, bends, knees, sharp-edged entrance, etc., and / that for surface fric- tion. Therefore we have from which the velocity is obtained. If Q be the required quantity, and S the section, then 2. Water flows from a tank through a 1 in. vertical pipe. Find the head in the tank so that the velocity of discharge jnay be the same for every length, taking into account the resistance of sharp-edged entrance as well as of surface fric- tion. 868 RESISTANCES. [235.] Let I be the length of the pipe, and H the head above the orilice ; then or H 2y But the conditions require that the term containing I shall disappear; hence and where v is independent of I as required. Then and d 4 X. 005(1+ I) = 3.7 feet. (1 + *) 3. "Water flows from a tank through a uniform sloping pipe of diameter d. Taking into account the resistance at entrance, show that the water will flow with the same velocity what- ever be the length of pipe, if the pipe slopes at an angle, sin ^ = — (^^72^) ^^ ^0 ^^^^S *^^ ^^^^ ^^ *^® *^^^ ^^^^® the entrance. [236.] . GASES. 369 We have, as before, We will suppose so small that the resistance to sharp- edged entrance may betaken asi^= i. We have/* = .005 — -7— , and 77i = 7- . Hence we have which may be regarded as an equation of identity in I, There* fore equating like terms, or Also 2g 1.5 em d = ^ — V 2^ d M{d+l)rr GASES. 236. The density of an ideal, incompressible fluid is inde- ])endent of the pressure to which it is subjected, and depend- ent only upon its constitution ; but the density of a compress- ible fluid is dependent upon both its constitution and external pressure. Thus, if water were strictly non-compressible, its density at all depths in the ocean would be the same as at the surface ; but the atmosphere, beino^ a gas, diminishes in den- sity as we ascend, since the weight of the atmosphere above 24 S70 GASES. [237.3 any point is less the higher the point. The laws of the pressure of fluids given in Articles 174 to 178 inclusive are applicable to gases. But if the pressure varies, the density and temperature both vary according to laws which are deter- mined by experiment. 237. Boyle's (or Mariotte^s) Law. — According to the experiments of Boyle and Mariotte, the volume of a gas at umform temperature varies inversely as the pressure to which it is subjected. Hence, if v^ be a known initial volume of gas, and j?o the initial pressure per unit of area to which the gas is subjected — being the pressure upon its bounding sur- face, or at any point within it, for which latter reason it is also called the tension of the gas — and v and p any other con- temporaneous volume and pressure ; then we have jpv = PqVq — constant = m, (414) which, if p and v vary simultaneously, is the equation of an equilateral hyperbola referred \ to its asymptotes. In testing V this law experimentally, it is j \ necessary, after compressing or j \. dilating the gas, to make ihe j \v5 temperature of the gas the ^° ''^>^^ same as the initial temperature ; ViP^ i^ in the former case cooling, and j ! in the latter heating it. The ._^;l_ I ^ Q — X law is found to be very nearly^ " '^° ^ but not exactly, correct for air and other known gases, the Fw. 182. ^ 1. • i.1, agreement being nearer the more perfect the gas, from which it is inferred that it cor-' rectly represents the law for the ideally perfect gas. Since the density, 8, varies inversely as the volume, we have, for imiform temperature, 6v = 8qVq = Gomta/nt, (415) and 6po=^dop. (416) [288.] GASES. 871 238. To FIND THE TENSION AT ANY POINT OF A COLUMN OF HAS OF UNIFORM TEMPERATUKE. — On Eccount of the Weight of the gas and its compressibility, its density will vary as some function of the height. Conceive a prismatic column of gas whose base is unity, and at the lower base j?o the tension, 6q the density, Wq the weight of a unit of volume, and at height z let the corresponding quantities be p, d, w. Then will the pressure on the lower base equal the pressure at the height z added to the weight of the prism of gas of that height ; or, and differentiating, dp = — wdsy (417) which is independent of the law of pressure and of the total head. According to Mariotte's law, ^■^o::~:^::p:p>, (418) 4nd considering gravity as uniform, w = -V. (419) Substituting and reducing, I> jPo integrating between the limits^ and^o, s and 0, gives 1 P ^0 ^Fo Po ^•z .\p=p^ ^» . (420) S72 GASES. Similarly, w = Woe P' ; 6= S^e ^» . The weight of a prism of gas of height h, will be W = wd3 = Wq e ih dz — Pq Jo 1 — e i>o [239, 240] (421) If A = 00 , W = 2Jo, as it should. 239. Numerical values. — The mean pressure of the atmosphere at the level of the sea is j?o = l^.T pounds per square inch, or 2116.8 pounds per square foot. The weight of a cubic foot of pure dry air under the press- ure of 14.7 pounds per square inch, and at the temperature of melting, ice (32° F.), Wq = 0.0S072S pound avoirdupois. (^23) If the atmosphere were pure, dry, and uniform, and of the density as at present at the level of the sea where the tem- perature is 32° F., the height would be b:= 2116.8 0.080728 s = 26,221 feet ; (424) or the height of one atmospliere of uniform density is nearly 5 miles; but as the atmosphere does not fulhl these condi- tions, and being a little heavier, the heiglit is a little less than this vahie. At the level of the sea the barometer stands at 29.92 inches nearly. At 5,000 feet above " " 247 At 10,000 feet (Mt. ^tna) << 20.5 At 15,000 feet (Mt. Blanc) ti 16.9 At 3 miles ** it 16.4 At 5 miles *' (4 8.9 240. To find the tension at extreme heights when the variation of gravity is considered. l^.J GASES. 373 From equation (418) we have y = ^-^p. (425) Let ^ be the radius of the earth, s any distance above the earth, then and (417), (425), (425a), give dp _ _Wo E^ , 7- 7o("^T^^ ' integrating between the limits of p and j^o? ^ and 0, p _ Wq Bz ^ and similarly, log' -D — ^Po PoR -^ Wq Rz V^R ^ Z :^. S^e ^o-^^* (425 J) Assuming the density of air at the earth Sq = f^j^ of a ])Ound per cubic foot, as it is very nearly, and Ji = 20,860,000 feet, and equation (4255) reduces to ^ = Thm ""*'. (425c) If 2 = 00 , we have '^ = 400x(10r ' ^'^^^ which is the limit of the density. If gravity be considered uniform, we would have for the height equal to the radius of * A more cumbersome demonstration of this formula may be made by means of Prop, xxii., B. II., of Newton's Principia. 374 GASES. [241.1 the earth, z — 20,860,000, and (421) and (418) reduce to the same as (425d). 241. Heights may be determined approximately by the pressure of the atmosphere. If its temperature were uniform and free from currents, we would have from equation (420), . = =^logi^»= 26,221 log. J, where \ and h are the readings of a barometer at the lower and higher stations respectively. Adapting this to the use of common logarithms, we have z = 26,221 X 2.30258 log — ^ ' =60,735 log ?y?. But this coefficient is known to be too large, and practice shows that 8 =60,346 log ?^ (426) gives better results. If % be the height of another station, then OQ QQ z, = 60,345 loa- ^^ ; from which subtracting the former, we have 2i -z = A = 60,345 log I- . (427) This result, however, requires to be corrected for the effect of the temperature on the mercurial column, the effect of the variation of gravity due to great heights, change of gravity due to latitude, the reduction of observations when not simul- taneous, the effect upon the mercurial column due to the ] OASB& 376 attraction of the mountain on which observations are made, all of which are well discussed by Poisson in his Traite de Mecanique, \i T = the temp'ture of attached thermometer at lower station, T' = " " " upper " t = " detached " lower " t' = " '' « upper " q) = the latitude of the place ; then, ^ - ^^'^^^ 1 - 0.00:^551 cos 2^ "" '"^^ S • I + 0.0001001 (r-rj *"'"'• ^^^^''^ Observations should be taken at both stations siniultane- onsly, in calm weather; but when this cannot be done, two observations may be made at one station at different hours, and one at the other at a time midway between, and the mean of the tbrnier be used as one observation. 242. GrAY LussAc's (or Charles') Law. — It was found by these men by direct experiment, that the increase of tension in a fixed volume of gas is directly proportional to the increase of temperature. I^i^o, to, Vq, be respectively the initial pressure, temperature, and volume of a gas, and p, t, v, any other corresponding contemporaneous values, and /3 a coefficient of expansion, then we have, the pressure being constant, v = Vo{l-h/3{t--to)), (428) .. the volume being constant, p=pil+fi{t-t^)). (429) From equation (428), 376 GASES. [243.] In practice, the initial temperature is taken as that of melt- ing ice, 32° F., or 0° C, and the initial pressure that of one atmosphere, or 147 lbs. per square inch, or 29.92 inches of mercury (760 millimeters). The initial volume may be of any convenient value. For these values it has been found that for carbonic acid, /3 = 0.003,7099 for 1° C, for air, /? = 0.0036706 " for hydrogen, /3 = 0.0036613 " From these values it appears that the more perfect the gas, the less will be the coefficient of expansion ; and Rankine concluded, in accordance with the theory of molecular vortices, that the limiting value is P = .000365 = g-f4- for 1° C. = 0.0020275 = — |- for V F., (431) which values are now used for the ideally perfect gas^ and aro sufficiently accurate for air and several other of the more perfect gases. 243. Absolute zero. — Equation (429) becomes, for ?5o-0°C., i>=i>o(l+2k)' (^^^) in which, if t—— 274°, jp becomes zero, and a perfect gas would be destitute of tension. Similarly, according to (428), the volume would vanish. In Fahrenheit's scale 4 is 32°, and equation (429) becomes ^ - 32 H^-^iiif)' (^^^) in which, if ^ = — 461.2, p becomes zero as before. Accord- ing to the modern theory of heat, at this temperature the molecules of the gas would be at perfect rest. It is the point of absolute deprivation of heat, and is called the zero of abso- 1244.J GASES. 877 lute temperature. No such temperature can even be approxi- mated bj any known process, tlie lowest recorded temperature obtained by artificial means being -140° C. (-284:° F.); but an extension of the law of uniform expansion leads to this result. Temperatures reckoned from the absolute zero are called dbsolute terrvperatures^ and are especially useful in simplifying many formulas in regard to heat. If t be the temperature on Fahrenheit's scale, and r the same tempera- ture on the absolute scale, Tq = 493.2, the temperature of melting ice, then r = 461°.2 -f 2f = To - 32° + 25, and equation (433) becomes jp=i>of. (434) If w be the weight of a cubic foot of dry air at pressure p pounds per square inch, and at absolute temperature r, then w = WQi^ ^; (435) and if w' be the weight at temperature r' and tension /?', then w'=w^'^,^, 6=do^.^; (436) and since for gravity uniform, the volume varies inversely as the weight, v = rof -f. (437) I Since the mechanical properties of gases, whether at rest or in motion, involve the property of heat, we now consider some of the abstract properties of the latter. 244. Heat is a form of energy. It is not force, since force is only one of the elements producing energy. See Articles 25, 26, 151. It is believed to be a certain manifesta- 878 GASES. [245, 246] tion of the motion of the particles of a body.^ The following principles, the result of observation and experience, have become established. 1. Heat may be transformed into external work ; and con- versely, work, nnder certain conditions, may be transformed into heat. 2. Heat cannot be transferred from a 'body of a lower to one of a higher temperature except by the aid of a machine and the expenditure of mechanical energy. 245. The thermal unit is the amount of heat energy necessary to raise a unit of weight of ice-cold water one degree on the therm ometric scale. The English thermal unit is the amount of heat energy necessary to raise one pound of water from the temperature of °32 F. to 33° F. ; the value of which, as found by Joule, is the heat produced by friction in bringing a body weighing one pound to rest after falling 772 feet in a vacuum.f Or, more briefly, the English thermal unit equals 772 foot-pounds of work. This is called Joule's equivalent, and is usually represented by J, The equivalent in French units is the heat energy neces- sary to raise one kilogramme of water from 0° 0. to 1° C, and equals 424 kilogramme-metres. The amount of heat energy necessary to raise the tempera- ture 1° is not the same from all temperatures, although for ordinary mechanical purposes it is considered constant for the same substance. 246. Specific heat is the heat necessary to raise a unit of weight of any substance one degree on the thermometric scale, the thermal unit being unity. Strictly speaking, sjpecific heat is a ratio, being the quotient obtained by dividing the quantity of heat required to raise the temperature of one pound of the substance one degree by the quantity required to raise the temperature of one pound of water the same amount. * Elementary Mechanics, p. 69. Stewart on Heat. f Elementary Mechanics, 1^^. 11, 12,. '■. [247,248.] GASES. 379 The specific heat under constant pressure of gases, is the Bpecitic heat determined when the gas is permitted to expand under the constant pressure. We denote it by c^. Tlie specific heat under constant volume is the specific heat determined when the volunie of the gas remains fixed. We denote it by Cp. The former always exceeds the latter, for in addition to: increasing the molecular motion of the gas, external work is; done against the pressure of the air, expanding it al)out 0.0020275 of its volume for each degree, as shown in equation (431). If the same quantity of heat will raise one pound of water m degrees, and one pound of any other substance n degrees, both under constaut pressure, then m ^ n The following are the values of <?p, c^, and Cp -^ c^ for a few gases: Gas. Cp Cv cp -hCv Air 0.238 0.169 1.408 Oxygen 0.218 0.156 1.400 Hydrogen 3.405 2.410 1.412 Steam 0.480 0.370 1.297 247. The dynamical specific heat is the specific heat ex- pressed in foot-pounds of work. If IT^ he the dynamical specific heat under constant volume, and ITp that under con- stant pressure, then K„ = Jc^, Kp = Jcp', (438) wliere J = 772 foot-pounds, or 424 kilogramme-metres. 248. ADiABA'no curve. — Let q be the heat energy neces- sary to change the temperature of a unit of weight of a gas i degrees, the pressure beings, and density d. Then .7=/(7,.ry.r). (439) Differentiating,^ being constant, and reducing by the aid of (436), gives 380 GASES. [248.J and differentiating, considering 6 as constant, /^\^d^,d^^p.<^^ (441) Wr/fi dr dj) t djf> Letting r = ^^ (^^2) ^r y being considered constant, (440) and (441) give in which r is involved only implicitly. Eliminating r from (439) by means of (434), and q=f{P'^\ (444) the total differential of which is do — dp —pydS in which substitute -^ from (443), and we find 1-7 6p ■J yS dq y_ ^=±^-dp ^ -f{4)<4}- ...^=(5V(^); (445) where/* is an arbitrary function, that part of which depend- ing upon -j^ is implicitly a function of ^ and d, as shown hj [248.] GASES. 381 (444), and the other part, S -r- jfT , is explicitly a function of the same quantities. The product of the two parts is some function of the same quantities expressed in terms of the ratio ^ -^ S ; the entire transformation being made so that an integral may be obtained, though it be in a functional form. 7^ represents some other arbitrary function, and q) the inverse of it. In the particular case where q remains constanU as it would when a gas is compressed or dilated in a close vessel imper- vious to the transmission of heat through its walls, although the indicated temperature would change, we would have for another pressure ^oi ^"d density (^o? po = ^l<pi<i) ; and eliminating cpi/j) between these equations, we have 2>6Z=j>,6y\ (446) and since the volumes will be inversely as the densities, we also have pv"^ = p^^v^ = constant, C^^'^) and from (436), (446), (447), we have r.= ©'"'=© '-©'"'=©'"'■ '"»' Fig. 183. Equation (447) is the sim- plest form of the equation of the Adiabat/lc carve, or "curve of no transmission" (from the Greek a, not, and 6ta/3aivstv^ to pass through), and is represented by the curve AJ3. If an isothermal line pass through the point A, where the volume of the gas is 1 and pressure pi, it 382 GASES. f34».i will pass above the adiabatic for values of v > 1, and under it for values of v < 1. E: XAMPLE8. 1. If a given volume of air has a temperature of 85° F.^ what will be its temperature when dilated to double the vol- ume, performing work, but receiving no heat ? We have, from equation (MS), /2^y-->»^_ 4f)1.2 + 85 ^ \vJ ~ 461.2 + t ' .'. t= - 50" F. 2. If a prism of air having a tension of IJ atmospheres at a temperature of 88° F. expands adiabatically to a tension of one atmosphere, required the final temperature. We have /I y-^^"^__ 4f^i.2 + t ^ Ui/ ~~ 549.2 ' .-. t = 24°.35 F. 3. Required the ratio of the volumes before and after ex- pansion in the preceding example. Ans. 1.335. 4. If air be compressed adiabatically from a tension of 15 pounds at 50° F. to that of 90 pounds, required the final tem- perature. 5. If air at 60° F. and six atmospheres expands adiabati- cally to one atmosphere, required the final temperature. 249. Velocity of a wave in an elastic medium.* Assume * The general problem of wave propagation has received the attention of several of the most eminent mathematicians since the days of Newton, and many problems have been solved in a satisfactory manner. The simple method of Newton, Principia, Prob's XLIII.-L, B. II., has not been excelled, and the definite theoretical result obtained is quoted to the present day, although the effect of heat upon the velocity of sound was not then known. La Place, in the Mecanique Celeste, tomes II. arid V., has treated of the oscil- lations of the sea and atmosphere ; Lagrange, in the Mecanique Analytique, [249.] GASES. 383 that the medium is confined in a prismatic tube of section unity, jFthe coefficient of elasticity for compression, p a force which will produce a compression dy in a length dx^ theu from definition we have The lamina ^a? will be urged forward — or backward — ^by tii« difference of the elastic forces on opposite sides of it, and as the quantities are infinitesimal, this difference will be dp ; or i>- = dr^Ep^ dx Let D be the density of the lamina, then its mass will be M = Ddx^ and we have from equation (21), page 18, d^ dx ' or, df~Dd^' ^^^^ which is a partial differential equation of the motion of any d^y lamina. Let E-r-D = o^, and adding a -7-^ to both members, we have dt \dt dxj" dx \di dx)' tome II. , has discussed the problem of the movement of a heavy liquid in a very long canal ; M. Navier published a M^moire on the flow of elastic fluids in pipes, in the Acndemie des Sciences, tome IX. ; and M. Poisson wrote several Memoirs on the propagation of wave movements in an elastic medium, and the theory of sound, for which see Journal de VKcole Poly' technique, 14th chapter, and of the AcadSmie of Sciences, tomes TT and X. These eminent mathematicians established the basis of all the analysis for the solution of the problem. More recently we have M. Lamp's Lemons s^ir r P^asticifi des Corpf< solides, and Lord liayleigh's Treatise on Sound, both of which are works of great merit. 584 GASES. [249.] Let then (S)'% <-) where the parenthesis indicates a partial differential coefficient and and equations (449), (451), (452), give fdV\_ (dV\ \dtJ~''\dx)' The total differential of V =/{x, t) is by substituting (452), and integrating, F=i^(.+«*)=|+.|, (453) where i^is any arbitrary function. Similarly, subtracting a -r-^. from (449), F'=/(«-«0 =§-«§• (454) f249.J GASEa 885 Adding and subtracting (453) and (454), and we have the respective equations ^ = iF(x + at) + if{x - (U)y But ■■•■"= (I) *+(!)* and substituting from above, gives di/ = -- F{x + at)d{x + at) - ^ f(x — at)d{x — at) ; 2ia Aa integrating, y = ip(x + at)+ cp{x — at), (455) where tp and (p are any arbitrary functions whatever. Their character and initial values must be determined from the con- ditions of the problem. The equation represents a wave both from and towards the origin. If the wave be from the origin only, the function may be suppressed, and we have y = tp{x + at), (456) and differentiating, (I) = n^ + '^ which is the rate of dilation (the expansion or contractiou of a prism of the air), and (J) = « • <i>'(<^ + <^i 386 GASES. [249.1 which is the velocity of a particle, and dividing the latter by the former, J = «, (457) which is the velocity of the wave ; hence, ^=«=j/^. (458) D The elasticity of air equals its tension ; hence if jp be the pressure per square foot, w the weiglit of a cubic foot, and H the height of a homogeneous atmosphere, equation (-1-24), then \ w Vgll; (459) hence the velocity of sound should equal the velocity of a body falling through a height equal to one-half the height of a uniform atmosphere. This principle is applicable also to the vibration of elastic cords, and it is found that The velocity of vibration of an elastic cord equals the velocity of a body falling freely through a height equal to half the length of the same cord whose tveight icould equal the tension. Similarly, in water neither too shallow nor too deep. The velocity of waves on the sea equals the velocity of a hody falling freely through a height equal to half the dejpth of the sea. It has been assumed that ^and D remain constant in wave motion ; but it was long since known that the results given by equation (458) for gases did not agree with those found by experiment, and La Place showed that the elasticity was increased by the action of the wave due to compression. It 18 necessary, therefore, to consider that the expression is'cor- rect only for ultimate values ; or . = |/f. (460> [260, 251.] (iASES. 387 Since E^ wy, dF= ^—dw = ^-^dD, w w ...u=^=Vg^. (461) 250. To FIND THE VALUE UK ;/, we liave from equation (461), y=p.~^ (462) by means of which it may be found when the velocity of sound in a gas of given weigiit and tension is known. In this way the values of y have been found for a variety of sub- stances, a few of which are given in Article 246. It is con- sidered as constant for any given gas, and nearly constant for all the more perfect gases. 251. Kemarks. It is difficult to find the specific heat of a gas under constant volume by direct experiment, but by means of equations (462) and (442) it may be readily com- puted when the specific heat at constant pressure is known. In this way the values of Cp, given in Article 246, were deter- mined. If the temperature of air in its quiescent state is uniform, the tension will vary as the density, equation (416) ; hence p -r- w will be constant, and the velocity of sound iu any gas will be the same at all temperatures or densities, bearing in inind that the factor y appears on account of the condensation resulting from the transmission of the wave. But if the tem- perature; varies on account of external causes, we have, equa- tions (435) and (461), "=i/^' (*''> in which the velocity is made to depend upon the absolute temperature. If Tq = 49.3.'2, /,, — '•o=2<),22I feet, equation S88 GASES. [252.] (424), and g = 32| ; the velocity in dry air at any tempera- ture r, will be - y ^ ^ '493V ^^'^^ ^=1.089.6 |/Ifeet (464) Examples. 1. Eequired the velocity of sound in air at the temperature of 0° F. 2. Required the velocity of sound iu air at 95° F. 3. If the weight of a cubic foot of hydrogen at the tension of the air, 14.7 lbs. and 32° F. be o'o05592, required the velocity of sound in it at 80° F., y being 1.4. 4. Required the velocity of sound along a steel bar, the coefficient of elasticity, jo, being 29,000,000 pounds per square inch, and w = 486 pounds. „ =^40,000,000^x^l44;XM = 16,625 feet per second. The velocity of sound in solids depends much upon their homogeneity. Experiment shows that its velocity is from four to sixteen times as great as in air. The above equations are no more than approximately correct for solids. 252. Yelocity of discharge of gases through orifices. If the density and temperature remained uniform during flow, the law of discharge would be the same as for liquids, and the same formula would appl}', and we would have for the velocity, v^ — ^gh^ where h would be the reduced head, being the height of a prism of the gas of uniform density equal to the density at the orifice. If the temperature be uniform, the density will vary according to Mariotte's la^V, or p = ~ d. Let u be the velocity, then will tlie mass passing a transverse section of area unity in a unit of time be Su, and the difference in pressures on opposite sides of a section will p52 J GASEa 389 be &p^ which will equal the moinentum of the elementary mass, or rfp = — Swdu ; (465) in which the quantities have contrary signs, since the greater the velocity, the less the dili'ereuce of the pressures. Divid- ing by^, integrating, 6. 'K^wJ'"'-^'^' (*««) where p is the tension of tlie gas in the reservoir, ^j the ex- ternal pressure, and if the gas flows into tlie atmosphere it will be 14.7 lbs., ii^ the velocity of exit, and u the initial velocity in the receiver. If the receiver be large, and the orifice small, p may be considered constant and u zero, in which case we have -yi ^•I«sf,; (467) in which, if j?j= 0, % = oo , or the velocity of a gas into a vacuum would be intinite according to this hypothesis. If the temperature and density both vary, the effective head will vary. Letting 2 be positive downwards, we have, from equation (417), J w Following the method of Joule and Thomson, and consider- ing gravity as constant, w will be a function of p and y ; and substituting and reducing by means of equations (448), we have 2y m which ^ is the tension within the reservoir, and p2 that just outside ; then if r^ = 0, the flow will be into a vacuum, 890 GASES. [3b8, 264.J and according to (468) the velocity will be a inaximuin, the value of which will be "=i/^^-F^i-£-^ w ^ which is a/ — — ^ times the velocity of sound; and for air becomes V = 2,41'"^ 4/ -^ ^^^^ P^^ second ; and if the air be 32° F., then V = 2,413 feet per second, or about 2.2 times the velocity of sound in air. 253. The volume of gas flowing out per second will he Q = sv, (470) where s is the area of the contracted section. 254. The weight of gas flowing out per second will be, equations (448), (468), 470), ^«..y[*i.5s(i-^;)](sr.(«.) which is a maximum for ' (r^~T P, ( ^ V~^ /'^^V . which for air becomes -' = 0.8306, £^ = 0.527, ^ = 0.6345. ^1 Pi oi APPENDIX I SOLUTIONS OF PROBLEMS. PROBLEMS. CONSTRAINED MOTION. 1. A particle is placed at the extremity of the vertical minor axis of a smooth ellipse, and is just disturbed ; show that if it quit the ellipse at the end of the latus rectum the eccentricity must satisfy the equation e* + 6e* + 3e^ = 5. Let V be the velocity of the particle at the end of the latus rectum, then v' = 2gh=iiff(b~^^y (1) Also in equation (148), p. 193, we have p=^a(l-e^){l+e^% X=0, Y='-m^. Prom the equation to the ellipse, dx 1 ^ ■y- = , — for X = aSy ^« Vi + ^ hence, v« = ga{\- e') (1 + «')t 7/1=^ = 5^^ (1 - ^% (2) Equating (1) and (2), ^a ( Vl - e' -{X-e'))= ga (1 - e% or, 2a/1 - c' = (3 + e') (1 - e') ; hence, * e* + 5e* + 3e' = 5. 2. A particle slides down the upper -surface of a frictionless 394 PROBLEMS. wedge, the wedge being free to slide on a frictionless horizontal plane ; deter?nme the motions. Let W — the weight of tlie body, W'= the weight of the wedge, h = AC, b = AB, 8 = A CB, N = normal action between the body and wedge, R = the vercical reaction of the plane AB on which the wedge slides, a = DB the distance moved by the wedge, h — a = AD, A CD = a. Take the origin of coordinates at C, x positive to the right, and y positive vertically down. The forces acting on the body are N and W, and on the wedge N, W, and R. The angle 6^, equations (143), is ~ 6, and 6y = 90° 4- ; and the equations of motion become — i^ = If cos 90° + JVcos (- d) = iV^cos 6, g df ^ ^ 0) w ^_ ' dt' W^d^ g dt g dt W'd'x' f = TTsin 90° + Wsm {~ 6) z= W - JV^sin ff ; (2) W cos 90° + N cos (180 - d) + R cos 270° = -Nco^d. (3) A' ^ ^'' ^'^ ^^° ^ ^^^^ (180 - ^) + i^ sin 270 = ; (4) which is zero, since the plane is fixed, and hence there will bo no motion in that direction. CONSTRAINED MOTIONS 805 From equations (1) and (3) we have W — -- W'^' " dt-- " df' integrating. ^ dt - ^ dt' and again. Wx= - W'x' ; or. W(h-a)= W'a; («) and W'b (6) Let 8 be the distance from C down the plane. then y = s cos 6, X = s sm — jy; differentiating. reducing d'y = cos 6 d^s, d^x— sin 6 d^s — J d^y. = (sin 6' — 7- cos ^jrf"*; which, in (1) and (2), gives 8 f/N cog '' W Urn <^ - I cos e\ dr8 _ gN cos 6 _ TT — iV^ sin ^ dP~ _/ . . a A"~ >fco8^ ^* ^^^ Solving, ffTf-sin^ ~ TF' sin' e + (W ■¥ W) cos» ^' ^ and 37" is constant. 3^6 PKOBLEMS. Integrating (1), (2), and (3), gives W^ = Wv,= g]Vco8 e . t, (9> W^ = Wv^ = g(W-N8me)ty (10) W'^= W'vJ =-gNco%d 't\ (11) and integrating again, gives Wx =igiVGOs d . t\ (12) Wy =ig{W- J^sin d) t\ (13) TrV= - i^i\rcos e . t\ (14) Making x — h — a\\i (12), a;' = — a in (14), and we find equa- tions (5) and (6) as before. Also make y = h va (13), and find JVas before. Then equations (8) and (13) give ^ ^ if ^^ / J f COS^ ^ + T;r^^TTr/ sin^ 6>^ (14') I ^ cos^ /? \ W + W J ^ ' Integrating (1) and (2) in reference to x and y, and we have for the velocity with which the particle reaches the foot of the plane. v^^/v^ + v^ =^n\x ^^'-'"'^ 1* (15) L (Tr+Tf')[(Tr+F')cos^^+Tr'sin^^]iJ ^ ^ From (11) find ,__ /«-^r TTTf^sin B "|i V, - V ^^^ ^^ p^ ^ ^^^ ^ _^ ^,j ^^g, ^ ^ ^, g.^, ^J . (16) The velocity with which the particle reaches the horizontal CONSTRAINED MOTION. 397 plane is readily found by the principles of energy. Thus the work done by gravity equals W7i, Hence we haye W W WJi = i~v'-{-i—v''; (17) ff 9 which, reduced by the aid of (16), gives (15), If JT' = oo, we have JV^= TFsin d, (18) V = v^, (20) v' =0; (21) which are the formulas for the motion of a particle down a smooth ^a;e^ plane. For the inclination of the path of the particle, we have A comparison of (15) and (19) shows that the particle acquires a less velocity when the wedge is free than when it is fixed. Questions.— 1. Show that whatever be the relative weights of the body and wedge, the common centre of gravity of the two masses will remain in the same vertical line during the motion down the wedge. (See Art. 149.) 2. Is the principle of the conservation of energy, Article 151, illustrated by this example ? 3. What is the relation between W and W, that will give a minimum velocity for the body "FT? (Eq. (15).) 4. Given W, equation (15), is it possible to assign sucb a value to W or to ^, or to both, so that v will be iV^ff^f iV'^gh or — '\/2gh ? 398 PKOBLEMS. 5. Is there an algebraic minimum to the normal pressure, equation (8) — or an algebraic maximum ? 3. A homogeneous vylijider rolls dow7i the upper surface of a wedge without slipping, the wedge heing free to move on a fric- tionless horizontal plane ; required the motion. The problem may be generalized, providing the successive elements of contact of the body and plane shall, be in the succes- sive elements of a cir- cular cylinder circum- scribing the body. The centre of gravity will move in -a vertical plane, and the body may be a cylinder, a screw, a coil of wire, a sphere, an ellipsoid of revolution, a volume of circles of contact with the centre of gravity between, and other forms. Rolling may be secured by a string (or flexible band) wound around the body, one end being secured at the upper end of the wedge, at G, and the othor to the body. Let W = weight of the body, ^ its radius of gyration, TF'= weight of the wedge, JV— normal reaction be- tween the body and plane, T the tangential action producing rotation, DB = z, AB = h, AD = h-z, AG=h, ACB=e; origin Sit C ; x positive to the right, and y positive downward, r the radius of the circle of contact. The forces acting on the body are gravity, W, acting vertically downward, the normal reaction, JV, acting directly away from the plane, and friction, or tension of the string T — whichever produces the rotation — acting upward along the plane. For the equations of the motion of the centre of the body we have (167), revolution having two Wd'x 9 ^ = If cos 90° + iV^cos (- 19) + J^cos (270° - d), (i) Wd^ g df ]f sin 90° + iV^sin (- 6) + 7^ sin (270° - 6). (2) CONSTRAINED MOTION. 899 The forces acting upon the wedge are its weight, W\ acting ■vertically downward, the normal reaction, jV', of the horizontal plane acting vertically upward, the normal reaction, N^ between the body and wedge acting downward, and the pull, T, acting along the upper surface of the plane downward. Hence we have _ ^ = PF' cos 90° + N' cos 270° + iV^cos (180° - d) + Tcos (90" - d). (3) The origin of moments being taken at the centre of the body, we have, equation (152), for the moment of rotation of the body. (4) (S) These equations may be reduced to — ^=iV^cos^- Tsin d 9 df E^=W- JSTsm 6 - Tcos 6 9 dt' — ^'= _ jvrcos6'+ Tsin 6 9 dV 9 dV Integrating twice, assuming JVand T constant, as they are, — x I =—{h-z)^\{NQo^e- r sin B) i\ (6) — y\ = y h = ^(W - N sm e - T COS 6) t\ (7) a;T = - — z=^(- JVcos ^ + Tsin 6) i\ (8) (10) 9 W 400 PROBLEMS. When the body rolls completely down the wedge, we hare rcp=CB=: ^W + ¥=a (say) (11) which in the preceding equation gives which reduces the preceding equations to W(i -z) = yN cos ^ . f - Z_^!^_?!2_?, (13) Wh = ig{W-]V^sme)f- ^^'"^f^^ ^ (14) W z = iffJVcos • f ^ . (16) Subtracting (15) from (13), gives Wb W+ W I 9 (16) ••• ^-^^ w+ W" ^^^^ From (13) and (17), r WW'b W¥a^md -] 2 _A , . from (14), "I a Jg(W-Nsia0) t'=[wh+''^^''"''"^' L r' W- Nem e (^*y) ' .:N= ^^ B ■\A sin e CONSTRAINED MOTION. 401 y^ A sin 6 y=3 W 2 W'k'a Wk' cos W ^^ . ^ '/•' sin' 6 —, -— + r' cos'' 6 + k* n + \\ _ r^ vyco8*6> + B'rY-4: Wk' ar'gcose{B -\-A8md) + 4 fK'-'/tr'g'^i V 4 r*r(/y + A sin ^) 7 ^- = j/5^ W / 2Wk'a-r'gA \ A sin 6 \ r'W / The entire work done by gravity is Wh, and this equals the energy of the body due to the velocity of the centre, plus the energy due to the rotation of the body, plus the energy due to the motion of the wedge ; or Wh = j^Mv"- + \Mk'cD^ 4- \M'v'\ 4. A 2)cirticle slides down a frictionless arc of a fixed vertical circle ; required the motion. The forces acting on the particle will be gravity, W, and the normal action, JV, between the body and arc acting normally outward, and will be the difference between the nor- mal component of the weight and centrifugal force. Using polar co- ordinates, origin at the centre, 6 the angle between r, the radius of the arc, and a vertical diameter, and resolving normally and tangentially, we have W d^ 9 "* df W sin 6, ©■=- W fd0\' — r 9 cos d — N, (1) m 402 PROBLEMS. Integrating (1) gives ^ '^dt j= -2gcose\=2g{l- cos 6), (3) which is independent of the n^ass. In (2) this gives ]V={3 cosO -2)W. (4) At the point where the body leaves the arc iV= 0, for which condition (4) gives cos ^ = f ; (5^ .-. ^ = 48° 11' +. The velocity at this point is, equation (3), dO ,^- r dt - ^ti/'-- (6) The subsequent motion will be that of a projectile, having the initial velocity of (6) and angle of depression of 48° 11' +. The normal pressure, N, varies inversely as 6, as shown by (4). To find the time of the contact, we have from (3), 'Vill-svi dd cos d =/»-'°«'""']l-.; The superior limit makes t = — oo, which shows that the particle will not start from rest at the highest point. It will be necessary either to give it an initial velocity, or place it at a finite distance from the highest point. 5. A lodij rolls down the arc of a fixed circle without slidiiig, starting at a point indefinitely near the highest point of the arc ; required the motion. CONSTRAINED MOTION. 408 Let be the centre of the arc GB, C the centre of the rolling body, W = weight of the body, R = OD, r = DC, the radius of the circle of contact, k = the principal radius of gyra- tion in reference to a hori- zontal axis through C, 6 = BOD, Kesolving tangen- tial ly and. normally, we have for the motion of the cen- tre C, — {R + r) ^ = r sin (? + i\^sin 180° + 7^ sin 270% — (R + r)(^ = Wco8 e + iV^cos 180° + ^cos 270% or W{R + 0(^'= ^> cos d - Ng. And for rotation we have (152), g dt' ' (1) (8) (8) where (p is the angle which some fixed line CE of the body makes with the vertical ; tnus cp will be zero when 6 is zero. Since arc BD = arc BD, and DCW= 6, DCE = <p - ^, we have r(cp- 6) = R6; .-. (p r Solving these equations gives T = }f+r' W sin e. (4) (5) 404 PROBLEMS. iV^=: Is COS ^ + -p^fy, (1 - COS ^) - 2 r. (6) When equation (6) becomes zero the body leaves the arc. Making ]^=zO, we find If the body be a sphere we have P = ^r\ and cos ^ = il ; .-. ^ =: 53° 58'. To find the velocity of the centre of the sphere when it leaves the arc, substitute J'from (5) in (1) and we find (^R + r)^ = ysmd, (8) and integrating and reducing, we find (-^ + *■) f = V^i/(l-co,ff){R + r), (9) which is the velocity at any distance from the vertical. Making cos 6 = 1^, we have for the required velocity, VHg{B + r). (10) To find the angular velocity of the sphere at the point where it leaves the arc, substitute dO found from (4) in (9) and make /cos ^ = 1^ ; we thus find From this point the sphere will continue with the uniform angular velocity given in (11), (the body having rolled by fric- tion, or at that point being freed from the string), and with the initial velocity given by (10), the centre moving as a projectile. The sphere will strike the horizontal plane tangent at the lowest CONSTRAINED MOTION. 4\jb point of the complete circular arc, when its centre is at a dis- tance r above that plane ; hence the sphere will be a projectile through the height li = Ecos cos-* \^-^ R —r = ^R-r. (12) The time of movement as a projectile will bo given by equa- tions («), p. 177, and will be given by the equation r- + 0.2254: V~RT~r • t = 0.104927? - 0.062177r. During contact it will rotate on its axis R ,10 ^r COS"^ — n 27rr 17 times, and as a projectile it will rotate Got ^. TT— times. tiiTt If the arc be not infinitely rough, let // he the coefficient of friction, rotation being caused entirely by friction ; required the motion for this condition. The friction will be ^N, and so long as 7* < f.iN the body will roll, but from the point where T = ^iN the body will slide on the arc, and the force producing rotation will be ^N. The first part of the motion will be given by equations (1), (2), and (3) ; but during the latter part /^JV must be substituted for T in (1) and (3), and the equations integrated again, the inferior limits for 6 and cp being the values found from (4), (5), (6), and T = piN, During this part of the motion, equation (4) will not be true, but instead thereof, if 5 = rep' be the total arc slipped over, we have Rd=r{cp- -^ cpf). (13) That there be no slipping we must have or from equations (5) and (6), jr^sin e < /,.^3 C08 e + j^, (1 - COS 6) -2J. (14) 406 PROBLEH& Equation (3) becomes f^'S = .irr. (15) and (1), W{R + r)^=Wg sin S - /liT^, (16) and (2) remains, W(R + r) (^y= Wg cos 6 - JV^^. (t'O Eliminating JV between (16) and (17) gives ^(-^ + *■) [S ~ ^ O T = ^^'^ ('''' ^ - ;i COS S), which, if it could be integrated, would give 6 =f{t), and then dd . -^ m (17) would give jY, which in (15) makes known, by inte- gration, cp = F {t) ; and these results combined with (13) would make known qj' in terms of t. The time of the movement and the superior limit of 6 during contact would be found by making N = \TL the value found for iV. If the body had a rolling motion only, it would leave the arc when 6 = 53° 58', as shown by (7) above, and if it slid off without friction, starting from the highest point, it would leave it at (9 = 48° 11', as shown by the preceding problem. If the body both slips and rolls, it will leave it at a point whose angular distance from the vertical is less than 54°, and greater than 48°. 6. Wiiat must be the radius of the rolling body so that it will touch the horizontal plane at the instant it leaves the arc, after having rolled from the highest point ? 7. What is the relation between the radius of the body and of the arc that the body shall roll twice on the arc from the highest point, before leaving it ? 8. What must be the relation between thie radius of the aro and that of the sphere, so that by rolling from the highest point of the arc to the horizontal plane through the lowest point of the circular arc the body will rotate once, or twice, or n times ? CONSTRAINED MOTION. 407 9. The base of a hemisphere rests on a horizontal plane, and a sphere rests at its liighest point ; if, from a slight disturbance, the sphere rolls off the hemisphere, how far from the cantre of the hemisphere will it strike the horizontal plane, and what will be the angular distance from the point where the sphere strikes the plane to the point on the sphere originally in contact with the hemisphere. — (Solution in TJie Mathematical Visitor y Jan., 3881, p. 191.) 10. One spliere rolls down another, the latter being free to roll on a horizontal plane ; required the motions. Let R be the radius of the lower sphere, M its mass, 6 thfe' angle through which it will have rolled in time / from a fixed vertical line ; r, m, d', corresponding values for the iippor sphere, (p the angle between the line of the centres and the vertical, T' the tangen- tial action between the spheres, T the friction on the horizontal plane, and N the normal action between the spheres. Take the origin of co- ordinates at the centre of the lower sphere, x horizontal and positive to the right, y vertically upward, and ^ the initial angle of q). For the sake of distinction, let x' and y' apply to the upper sphere. The figure represents the condition of the bodies at the end of time t. The tangential action between the spheres will, at first, act downward in reference to the lower sphere and upward in reference to the upper sphere. The normal action, N, will not produce rolling of the upper sphere, but will tend to produce foiling of the lower one in an opposite direction to that of the tangential action. Any modification of these assumptions du» to the motion will appear in the solution. For the lower sphere we have 408 PROBLEMS. fPor M^ =iV^cos(270° + ^)+r'cos(360°-^) +rco8l80^ = — li sin (p + T' COS q) ^ T df = MJc'%^,= TR ■\- T'R dv or|.¥ij|f=r+2" 5(1) and for the upper sphere, d'x' m df N mi cp— T' cos (p m -^ = xY cos ^ + ^' sm ^ — mg dt >; (8) and for the geometrical conditions, x=E6; dx= Rdd, (3) x'— x — {R + r)B>m.cp\ .\ dx' —dx—{R-\- r) cos (pdcp, (4) • y' =z (R + r) COS cp ; .'. dy'=—{R + r)^mcpdq), (5) Jt{cp-e-p) = r{e'-cp + ft) ; R{dcp-dd)^r(de'-d(p). (6) Eliminating N, T, and T' from equations (1) and (2), gives df • (^ Integrating once, the initial values being 0, gives (8) ^^dx dx' „ dd' *^4^ /7^ ?7Z/* dO from which we see that when -^ = ^r^^ • -rr * ?^^ ^«^^ if F=s 6?^ MR dt — mv ; or the moments are numerically equal, and the centres CONSTRAINED MOTION. 409 are moving in opposite directions, ( V and v being respectively the velocities of the centres). Eliminating dS, dd', and dx' by means of (3), (4), and (6), gives :i{M + m)^ = m{R + r){\- cos q>)^. (9) and |(Jf-hw)^ =K|-cos ^)+i(Jf+m)co8 cp]{R + r)^. (10) From (9) it appears that the motion of the centre of the lower sphere will he positive so long as cos cp is less than ^, or cp < 66"^ 26'. Should the spheres separate at a less angle, they will con- tinue to roll in the same direction. From (9) it appears that when cos ^ = f , the centre of the lower sphere will be at rest, in which case (3) shows that there will be no rotary motion ; in short, the lower sphere will then be at rest. Equation (10) shows that the motion of the upper sphere will be continually positive. By means of the equations given above, a complete solution may be found, giving the nor- mal reactions and angular velocities. 11. ^ uniform rod of length a, capable of making complete revolutions in a vertical plane about one extremity, is placed in a vertical position with its free end upward, and being slightly displaced, moves from rest ; find the time of revolving from an angle /3 to an angle 6. The equation of motion is. Art. 126, a'drd a . ^ *^3^ = 2^'"^^' 3 g dt' Multiplying both members by ^dd and integrating between the limits 6 and 0, we have 410 PROBLEMS. or .: t + G=(^y log tiuii9. Hence the time from /J to ^ is , /2a\4, tani* '=fej'°»ti4§- 12. ^ rod rests with one extremity on a smootJi plane and the other against a smooth vertical wall at an inclination a to the horizon. If it then slips down, show that it will leave the wall when its inclination is sin "'(f sin a)» Let the mass of the rod be m, its length 21, its inclinafcion to the horizon 6, and the coordinates of its centre of gravity x and y; the origin being such that for the time, t, considered, x — IqoqO and y = / sin 0, Let the horizontal and vertical reactions at the ends of the rod be JJ and F respectively. Then the equations of motion are d'^y ^d^i^m 6) ^ „ .^. d'x , d^ (cos ff) _. i'^^j^j = HI sin 8- VI cos $. (3) Multiply (1) by cos 0, (2) by - sin 0, and (3) by (1 -7- 1), and add the products. There results i^^^ = - '^9 cos e, (4) CONSTRAIl^D MOTION. 411 fid whence, since ;5t = and 6 = a when ^ = 0, -:^-^(sm«-8in^. The rod will leave the vertical wall when ») ^=-»az(sin^|f + C08S^^. = «. Substituting in this the values of -^ and -^ given by (4) and (5) we have ^ = sin-* (f sin a), (The Analyst, 1882, p. 193.) 13. An angular velocity having teen impressed on a heterO' geneous sphere, about an axis perpendicular to the vertical plane which contains its centre of gravity G, and geomelrical centre C, and jjassing through O, it is then placed on a smooth horizontal plane. Find the magnitude of the impressed angular velocity that G may rise into a point i)i the vertical line SCK through C, and there rest ; the angle GCS being a at the beginning of the motion, a the radius, and cp the required angular velocity. Draw the radius CGA, and from G drop the perpendicular GM to the plane. Let m = the mass of the sphere, k the radius of gyration of the sphere about an axis through G perpendicular to the plane containing C and G, R the mutual reaction of the sphere and the plane, SM = X, GM = y, CS = a, angle A GM = angle ACS= <p, and CG = c. Since there is no friction, we have for the motion of the centre of the sphere d'x - 0) 412 PROBLEMS. resolving forces vertically, '^^^R-fng, (2D and taking moments about G, m¥ -—- = — Ec sin (p, (B) <p being the angular motion of the sphere. We have y = a — ccos cp, whence Substituting in (2), ^ / . d^(p dcp* \ R =z m(c sm cp-^ + c cos cp-^ + gj. This in (3) gives by reduction, (c' sin' cp + k^) -^ + c^ sin cp cos q) -—■ = — eg sin <p, (4) Integrating, (c* sin^ q? + k')^=C-h %cg cos cp. (6) Let ^ = 0, when cp— a; -^ = go, and C = (c^ sin' a + ^'^)g9' — 2c^ cos ou Hence (5) becomes (c* sin'* ^ + ^') "^ = ^^' ®^^' « + P)(»2 + 2c?^ (cos 9> - cos a). (6) Now if the initial value oi cp = a, the terminal value =z7t — a, hei and when also -^ = ; then the left member of (6) becomes zero^ « _ ^cg cos c^ ~ & sin'* or + F * (7%e Analyst, July, 1882.) KINETIC ENERGY. 418 Kinetic Energy. 14. A tally mass rriy radius r, is shot with a velocity v into a perfectly hardy smooth tube of length a, radius r\ mass m', free to turn about its middle point, which is fixed, imparting to the tube a rotary motion ; if the ball just reaches the centre of the tube, required the angular velocity of the latter. In this case the kinetic energy of the ball and tube due to the rotary motion, will equal the kinetic energy of the ball before it enters the tube. Let k and k' be the radii of gyration of the ball and tube respectively, and &? the required angular velocity, then, equation (153), imv* = ^mk'oo^ + im'k'^aj^ ; G?' mv' mJc" 4- m'k'* ' If the tube be considered slender, we will have h* = 3*^' k^ = |r', hence, 0^ = also 60mv' %^mr^ + bm'a^ ' 15. A cone, m,ass m and vertical angle 2a, is perfectly free to move about its axis, and has a fine, perfectly smooth groove cut in its surface, making constant angle fi with the elements of the cone. A heavy particle, mass M, moves along the groove under the action of gravity, starting at a distance c from the vertex; required the angle through which the cone has turned when the particle is at a distance r from the vertex. Let p be any variable distance from the ver- tex, (p the angle through which the particle has moved, k the radius of gyration of the cone, and 6 the required angle. The geometrical relations give /) sin ar = radius of the cone at the place of the particle at time t, p sin adq) = the horizontal arc through which the particle moves in time dt, dp tan fi = the same arc ; 414 PROBLEMS. /. dp = p sin a cot ftdg), (1) The moment of the momentum imparted to the cone in an element of time, will be dt The angular advance of the particle will be dq) — dS, and the horizontal component of the momentum of the particle will be ,^ . fdo) — d6\ and the moment of the momentum will be M,^^.^.(^^-^). Since gravity has no horizontal component, the horizontal motions will be due to the action and reaction between the bodies, and these moments must be equal ; hence Eliminating dq) by means of (1) and (2), we have f ' 2 J/ sin^ apdp f,-. . ^ aio —n nr -1 • 9 = 2 sm «' cot add ; ,, , tan 6 , m¥ + Mr^ sin^ a • • , ^ = 4 — losr • ^ sin o' ° mk^ + Mc^ sin'* a 16. An elastic ring, mass m, natural radius «, modulus of elasticity f, is stretched around a cylinder ; the cylinder sud- denly vanishes ; find the time in which the ring will collapse to its natural length. 'nam 8f" T=^''- 17. A prismatic har in a vertical position rests on a pivot at its lower extremity ; it is slightly disturbed, required its kinetic energy when it will have rotated 180°. KINETIC ENERGT. We have, Prob. 19, p. 216, and Art. 107, 415 dt' W'\V and integrating 1 dff' S g ^ 2^ = 2 f^^^' ]> 2 ^ ff which in equation (153) gives i 1 7, ^^' TF7; that is, the energy is the same as if the bar had fallen freely through the vertical descent of the centre of gravity of the bar. IS, If a sphere f pivoted on a horizontal dia- meter as an axis without friction, oscillates about an external axis ; required the kinetic energy when vertically under the support. Let the body rotate about the axis of y, then will equations (177) be applicable, and we shall have AB = l,L^= 0, J/i = - Pf7sin (9, N^ = 0,Mg = W,x = isin 6, z = IcoB 6 ; .'. d'x= —I sin ddB* + I cos Od'S, d'z= -I cos ddff' - I sin Od'O ; and these, in equations (177), give df" l^sin^; which, integrated, gives f = fcos.];=f(l-cos^. ide" 2 The velocity of the centre of the sphere at the lowest point will be 416 PROBLEMS. ,dd and the kinetic energy of the mass when it rotates through 180* will be ! hence in this case also, the kinetic energy is the same as if the sphere had fallen freely through the height equal to the descent of the centre of gravity. 19. Suppose, in the preceding example, that the axis of the sphere he rigidly connected with the rod AB ; required the velocity. Here we have, as in example 19, p. 216, d'e gi ' n -'d¥ = rTk}''''^> which, compared with the preceding problem, shows that the angular velocity will he less at the lowest point when the sphere is rigidly connected with the rod AB, than when it is free to 7'oll on its own axis. The kinetic energy in this case, when it will have rotated from the highest to the lowest point, will be, equation (153), page 202, i-m{P + ir').^ = 2Wl; which is the same as in the preceding case. The time of vibra- tion will be greater in this case than in the preceding — equation (161). Queries. — 1. If the sphere in example 18 be free to rotate on its horizon- tal axis as a diameter while the entire mass rotates about an external axis, and it rotates through 180°, starting with no velocity from a point vertically over the fixed axis ; if, at the lowest point in its path its axis instantly becomes rigid with the bar AB, will it rise to the highest point ? MOMENT OF THE MOMENTUM 417 2. In the preceding example, will the time of describing the second pari Df the arc be the same as that of describing the first 180° ? 3. In example 18, if the sphere gradually melts away, will the Felocity or time of vibration be thereby affected ? 4. In example 19, if the sphere gradually melts away, will the time or velocity be thereby affected ? 5. In examples 18 and 19, which will produce the greater stress on the axis of suspension, the masses and arcs of vibration being the same in bath cases ? 6. At what points of the rotating masses must they strike a fixed obstacle so as to produce no shock on the fixed axis ? 7. If a spherical shell B be rigidly connected to the bar AB and filled with a frictionless fluid, would the time of vibration and the velocity at the lowest point remain the same if the fluid should suddenly freeze ? 8. If a spherical shell rigidly connected to a bar and filled with a friction- less fluid, be rotating about a vertical axis with a uniform velocity under the action of no forces ; should the fluid suddenly freeze, will the velocity of rotation remain the same ? Will the kinetic energy remain the same ? Conservation of Areas (Art. 150). Moment of the Mo- mentum (Art. 166). 20. A cylinder of ice, radius r, length l, revolves with a uni- form angular velocity Ci? ; if it he subject to no external force and melts, required the angular velocity of the resulting sphere. Neglecting the contraction due to melting, and the spheroidal form due to rotation, and letting gd' = the required angular velocity, F = f 7?* = the principal radius of gyration, we have. Article 166, m • \r*Go = m • f i?*<»'. 5 r* /. Gt> = To find R we have .-. a,=j^<«. i7tR'= Tir'l; , 5 /4 r\i 418 PROBLEMS. 21. If n spherical shells of infinitesimal thickness, mass m of each, radius r, m,ove in contact without friction, having their axes of rotation all in one plane, the angles between the axes and an assumed line being (3^, fi^, j3^, . . . . /?„, and having angular velocities coi, gd.2, go^, gd^, respectively, suddenly become solid ; required the position of the resultant axis, and the resultant angular velocity. Let 6 be the required angle and oo the required angular velocity; F = |r^ the principal radius of gyration before and after becoming solid ; then, since the moment of the momentum will be constant, we have, resolving parallel and perpendicular to the line of reference, WTW^ cos 6 ' oD—m'k\GD^ cos y^i -f- &?, cos /?2 + • • • • c»„ cos p^=mT<?A ; nmk* sin B • Go=mh\csDi sin /?i + gd-i sin /ig + • • • • &?„ sin fi^) =m]c^By where A and B are the values respectively of the parenthetical quantities. Reducing, we have GO VA' + B^ d = cos* ' — ==- = sm- 1 — . ^/A' + B'' VA' + B' 22. A prismatic bar rotating in free space suddenly snaps asunder at its centre ; required the subsequent motion. The body will rotate about its centre, and after separation each half will rotate about its own centre, and those centres will have the same uniform velocity that they had immediately pre- ceding separation. After separation there will be two independ- ent systems, still at the instant of separation the entire moment of the momentum will equal that of the original bar. Let I be the length of bar, m its mass, h^ its principal radius of gyration, cox its angular velocity, gj the required angular Telocity of each half after separation, v the velocity of each half, and h the principal radius of gyration of each half; then we have also MOMENT OF THE MOMENTUM. 419 which, substituted in the preceding, gives CO = ooi ; hence The angular velocity/ of each half after separation will he the same as that of the original bar, and the two halves will move in opposite directions with a uniform velocity. Since no forces are conceived to act upon the bar, the kinetic energy after sepa- ration will be the same as before ; hence we would have, equation (154), 2(imv') + imh'oD' = imh\oo\ ; or as before. Suppose that such a bar separates into n equal parts, what will be the subsequent motion ? Or if it suddenly melts, or dis- Bolves, will the elements partake of the rotary motion ? If two or more bodies having a rotary motion cohere, will the aggre- gate mass have a rotary motion ? Can rotary motion impart a motion of translation ? 23. If a bar rotating about one end suddenly snaps asunder, required the subsequent motion. 24. If a bar rotating about one end gradually melts away at the free end, will the angular velocity of the remaining part be chansrod ? o 25. A spherical shell of infinitesimal thickness, wass m, radius r, is filled with a frictionless fluid, mass m' ; the shell is rotating with an angular velocity ai when the fluid becomes solid arid rotates with the shell ; required the common angular velocity of the mass. Let Cfo' = the required angular velocity. The moment of the momentum of the shell when the included mass is a fluid, will 420 PBOBLEMS. be I mr*GDy and of the entire mass after it becomes one solid will be |mr^OL>' + m'k^oo' ; If k' = |r*, then If m' = 0, as it should. , 5m oo' = ■= T-o—, CO. 6m + om GO = CO, 26. A spherical shell, external radius r, infernal radius r^, mass m, filled toith a frictionless fluid, mass m', rolls on a per- fectly rough horizontal plane luith a velocity v ; the fluid freezes and rolls with the shell ; required the velocity v' of the common mass. Let GO be the angular velocity before freezing, and go' after. The moment of the momentum before freezing will be mk^oo f (m + m')v • r, and this will equal the moment of the momentum after freezing, hence mh^GD + (m + m')vr = mk^co' + m'klGo' + (m -f m,')v'r, also , 2r' - r\ V = VGo, v' = rGo', k\ = ^rl, *. v = b r' ~ rl' 2m(r^ — r?) + 5(m + m'){r^ - riy 2w (r^ — r\) + 2m V^ (/•' — rf) + 5 (m + m') (r^ — r?)r' If m' = 0, we have v' — v, as we should. The kinetic energy of the entire mass After freezing will be less than before ; and, generally, whenever the internal forces cause a relative change of parts or particles of the system, the kinetic energy may be changed. 27. A circle is revolving freely about a diameter with the angular velocity go, when a point in the extremity of the perpen- MOMENT OF THE MOMENTUM. 421 dicular diameter becomes suddenly fixed ; required the instan- taneous angular velocity go'. We have mk'i GO = mk-Qo' ; 1^2 /. GO =T,00=j-^G0=i00. 28. A cone revolves about its axis with an angular velocity go ; the altitude contracts, the volume remaining constant, required the resultant angular velocity. Let h be the original altitude, x any subsequent altitude, r the original radius, y the radius when the altitude is x ; then, the volume being constant, ^7ir'h = \7ty% the moment of inertia will be mJc^ = -^Tir^h, mk\ = ^ny'x ; but the moment of the momentum being constant, mk\ GO = mk^GOi ; kl r' X or the angular velocity will vary directly as the altitude of the cone. 29. One end of a fine, inextensible string is attached to a fixed point, and the other end to a point in the surface of a homo- geneous sphere, and the ends brought together, the centre of the sphere being in a horizontal through the ends of the string, and the slack string hanging vertically. The sphere is let fall and an angular velocity imparted to it at the same instant, the sphere winding up the string on the circumference of a great circle until it winds up all the slack, when it suddenly begins to ajicend, winding up the string, the sphere returning just to the 422 PROBLEMS. starting point. Required the initial angular velocity, the ten- sion of the string during the ascent of the sphere, the initial upward velocity of the centre of the sphere, and the time of move- ment. Let I = the leagtli of the string, r = radius of the sphere, m, = its mass, k = its radius of gyration, x = the length of the unwound portion of string at the end of descent, v = velocity at the end of descent, v' the velocity with which the sphere begins to ascend, go= the initial angular velocity, go' = the angular velocity with which the sphere begins to ascend, t^ = the time of descent, t^ the time of ascent, T= tension of the string during the ascent, and / the impulse communicated by the string to the sphere at the end of descent. The length of string wound up at end of descent is ' rcoti = raoA/ — , y 9 ti being the time of falling freely through the height x, '2x 7 A A^ ;^ X = 1 — rooA/ — whence -^j/&- « For the impulsive motion, CD — GO Ir mJc"* Eliminating /, But P = |r', and v = \/%gx, , I V -\- V — — . m — (go - go') = V -h »'. r ^ ' .•.«?' = (» ~|;(A/2^ + tO. (8) MOMENT OF THE MOMENTUM. 423 The upward motion, the origin being at tlie point where the centre begins to ascend, and the axis of y positive upwards. For the angular acceleration, or Also or mk'g^-rT, d^y _ rd'd df ~ dt' ' (5) Eliminating ^ and ^ from (3), (4), (5), T = \mg = ^ the d^d weight of the sphere. Eliminating T and -r^ from (3), (4), (5), -^ z=z — ^. Integrating, observing that when / = 0, dt -"^^ % = v'-^t (6) When ^ = 0, ^ = /f, ; ••• '--^gt^ (7) Integrating (6), observinir that when ^ = 0, y =0, y = v''t-^f. When y =^ X, t = t,, X = v% - ^gtl (8) 24 problems; From (7), (8), v' = Wgx, (9) . -i/^'- (10) From (5), dy dd dt ~^ dt' . 1 . v' = roo'y od' ~- V Y^a;. (11) Substituting these values of v' and go' in (2), x-^l{i6- V35). Substituting this yalue of :?: in (1) and (9), CO (a/35 -6)Vgl ^Vp(6- V35)]' From (10), v' = V[V-^^(6 - V35)], The whole time is (Problem by the Author in Mathejuatical Visitor, Jan., 1879.) ^ 30. If n concentric uniform spherical shells of infinitesimal thickness moving loithout friction ahout axes whose inclinations to three rectangular axes are a^, (3^, y^; a^, ft.^, y^, etc., with angular velocities gd^, ooz, etc., respectively, suddenly become one MOMENT OF THE MOMENTUM. 425 solid; required the resultant angular velocity and resultant axis. Let a, y5, y, be the direction-angles, co the resultant angular velocity, k the radius of gyration, and m the mass of each ; then. Art. 166, nmk*GD cos a = m]c\QOi cos ai + gd^ cos cn^ +, etc.) = A^ (say). nmk^oj cos /? = 7w^'(gl?i cos /?i + oo^ cos /^a 4- , etc.) = J?, nmk'^GD cos ;/ = w^'(g9i cos ;/i + oo.i cos ;/j +, etc.) = C; and from Coordinate Geometry, Art. 198, Eq. (3), cos* a + cos* P + cos* y = 1, Substituting from the preceding, we have A"" + B" + C^ = n'm'k'oo' ; _ V A' + B* 4- G^ ~ nmk^ This value in each of the preceding values gives A cos a = cos /? = cos y , ^ ^/A' + ^^ + a* If the motion be in the plane xy, we will have y^ = 90° = y^ — y^, etc.; hence y — 90°, and the third of the preceding equa- tions vanishes, and the case reduces to the problem given by the author in the Mathematical Visitor, Jan., 1882, p. 14. 31. A screw of Archimedes is free to turn about its axis placed vertically ; a particle placed at the upper end of the tube runs down through it ; determine the resultant angular velocity im- parted to the tube. VT + 3^ + 0'' B VA' + B'+G'' C 426 PROBLEMS. Let m = the mass of the particle, nm = the mass of the screw, a = radius of the screw, z = the vertical distance the particle has descended, and, 6 and cp — the angles through which the screw and particle have respectively revolved about the axis in the time t. The energy of both bodies equals that imparted by gravity, or ^^^ -df-^'^df-^ "^^^ ^ = ^^^'- (^) From the principle of Conservation of Areas, we have ^d(p ^dd „. dt dt ^ ' We also have the geometrical equation z = a{cp -{■ 6) tan a, (3) From (3), Substituting in (1), ,/ 1 dcp' ^sm'a dcp dd ^m^add' dd'X ^ ,^. \cos^ a dt^ cos^ a dt dt cos^ a dr dv J ^ ^ ' From (2), dq) _ dd U-'^di' Substituting in (5), 2w sin^ a sin' a \d6^ or. J 'n/ 2n sm' a sm' a \dfr _ Vcos'* a cos* a cos^ a J dv ^ df^ a\n + l)(n + sin' ^) Jf — ^^^ ^^^' ^' MOMENT OF THE MOMEIJTUM. 42? When -Tfj = CO, z = h. at 2gh cos' a l)(n + sin' a)' 32. A cube slides down an inclined plane with four of its edges horizontal. The middle point of its lowest edge comes in contact with a small fixed obstacle ; determine the limiting veloc- ity that the cube may be on the point of overturning. Let V be the velocity of the cube at the instant of impact, 2 J the length of one edge of the cube, k the radius of gyra- tion in reference to the edge at P, and ki the principal radius of gyration in reference to a parallel axis, gd the initial angular velocity, and m the mass of the cube. The moment of the momentum just before impact will be mv b. (1) The initial moment of the momentum after impact, vrill be, equations (155) and (123), and example 4, page 172, Qa - mk'GD = m(2b' + kl)GO = m(2b^ + f5«)(w = ^mb'oj. (2) Hence, Article 166, mvb = ^mb*cj ; .-. V = ^boj. (3) The cube will be on the point of overturning when the energy due to rotary velocity is just sufficient to raise the centre to a point vertically over P. The energy will be, equations (153) and (123), i2mr • 00* = ^mk* • cy» = imoo^b' = ^mb'co'. (4) The work of raising the centre to its highest point will be 428 PROBLEMS. mg l\/^\l — cos (i;r — p)\ which being equal to (4), we have («) <»' = f Va f- [1 and this substituted in (3) gives JJ cos (Jtt - /?)], 5V2 • hg\\ - cos (i-TT - fi)\ (6) (7) which is the required result. In this problem we may consider two impulses, one that of the momentum before impact ; the other that destroyed by the impact. It is now required to find the magnitude and direction of a single impulse, which applied at P, will produce the same effect. At first it seems that this impulse will be parallel to the plane and opposed to the direction of motion, but if the body were free, it would, in this case, rotate about its centre, equa- tions (168), in which case the corner at P would move perpen- dicularly to the diagonal through the centre, whereas, in the problem, this corner becomes instantaneously fixed. The result- ant impulse may be considered as the resultant of an impulse acting along the plane, and another acting at P perpendicular to the diagonal of the centre through that point. The angular velocity is, equation (3), 00 8 I (8) hence the actual velocity, w, of the centre is which is also the initial velocity with which the point P would move normally to the diagonal if the body were free. The plane and obstacle then, impose tlie two component velocities v and u^ the angle between which is' |;r ; hence the resultant velocity V will be MOMENT OF THE MOMENTUM. 429 F' = v' + w' + 2vu COS }7r, = «^* + A^'-K; .-. r==iV34t;. (9) To find the angle between V and v, we have u* = V^ + v^ — 2vV cos cp ; which is constant, as we might have anticipated from (9), since V varies directly as v. To find the energy lost by the impact, we have, for the kinetic energy before impact, equation (24), imv* ; (11) and for the initial energy after impact, equation (153), problem 4, page 172, equation (123), and (8) above = Amt'«, ' (13) which is independent of the dimensions of the cube, the mass remaining constant. This is only | of the energy in the cube before impact, hence | will have been removed and passed into heat. 33. A circular disc, radius r, mass m, rolling on a rough horizontal plane with a velocity v, impinges against an obstacle whose height is b j if there be no slipping on the obstacle^ required the velocity immediately after impact, the height to tvhich the centre of the disc will be raised if it does not pass over the obstacle, and the velocity at the highest point if it does pass over, and the velocity of approach that it may Just roll over the obstacle. The velocity immediately after impact may be found as in Article 145, and the other results as in the preceding problem. 480 PROBLEMS. 34. A uniform bar of infinitesimal section, length ?, mass m^ moving with a uniform velocity v, strikes an equal har at rest hut free to move ; if the two Mrs are mutually perpendicular, and the former moves in a path perpendicular to both, required the motions of each after impact, supposing the end of one is struck by the end of the other, both bodies being inelastic. What will be their subsequent motions if they rigidly adhere at the instant of impact ? Let OA be the position of the moving rod, OB that of the rod at rest, at the instant of impact ; OX the direction in which was moving, u the velocity with which the centre, G, of OA was moving before impact, v its velocity after impact in the direction OX, v' the velocity with which the centre D of the other rod begins to move ; Ic the radius of gyration of each rod about its centre, go, gd', their respective angular velocities about their centres ; Q the impulsive reaction at 0. Put il = a, we have mv -- mu — Q, m¥oD — Qa, mv' = Q. mJc^oo' — Qa, (1) (3) (3) (4) The velocity at of the moving rod after impact is v — «&? ; that of the other rod is v' + aoo'. Since the rods are inelastic, these velocities are equal, hence V — ao3 = v' + aoa\ These five equations readily give (5) Gi> = GO i^'^^- FRICTION. 431 If the rods rigidly adhere at the time of contact, the values of V, v'y 00, od\ found above, will act after contact on the connected rods. Join CD ; its middle point, G, will be the centre of grav- ity of the connected rods. Join OG, Let F= the velocity of G after impact in the direction OX, 2mV = mu ; .'. V= \u. The velocity of in the same direction is t; — aoo = \u. Hence 00 will move in the direction OX with a velocity = \u. iu • V — v' = ^u a' + ^' a' + Itf' Hence C and D will both begin to revolve about G with a a* velocity = ^u — ^ • The angular velocity of the system about ^'^ = ^ ^ ^ = Tf • ¥T¥ Substituting k' = K ; 3 u ' a = U; v' = ^Uf GD= gd' = —J- , and the angular velocity about OG, when the rods adhere, will be 3 w_ FRICTION. 35. Find the conditions that a hoop shall roll down an inclined plane without sliding, ^ being the coefficient of friction, and i the inclination of the plane. Let B be the angle through which the hoop has rolled at the time t, a the radius, m the mass, and F the friction. Then we have. Art. 1*^5, and therefore m —STT- > gm&mt — F. am sin i < 2^; 432 PROBLEMS. but F=gm}j. cos i ; therefore tan i < 2/i. 36. A sphere having a rotary motion about a horizontal axis is placed gently on a rough plane; determine its motion. Let W = the weight of the sphere, /* the coefficient of friction, T the tangen- ^^ tial action due to friction. There may be two cases : 1st, if there be slipping ; and 2d, if there be no slip- ping. In the former case T = jjlW. For the motion of the centre we have ^9 (1) and for the rotary motion, Tr = — mWt, or (2) Integrating, we have dx dt = /^gt -hCi X = \i^gt^ + c{t + Ci Jc*cp = — ipigrt^ + c't -V c'\. (3) For ^ = 0, we have x — ^, and ^ — /, Cj = 0, and c" = ; also for ^ — 0, -— — oa^, the initial angular velocity, .*. c'=^Gt7o; az FRICTION. 433 dx and -^ = 0, the initial velocity of the centre, and the corrected equations become ^^ A* d<p GOn (p = OOot 1 /Agrl* w dx The velocity of the point of contact will be plus ^ , and minus r -^ ; hence at ' dx da? . uqrH dt dt (5) and so long as this is finite the preceding equations will hold true, but when it reduces to zero the conditions change. To find when they change, muke each member of (5) equal to zero, and solve for t, which call t^ ; hence t^ GD^rk* jjg(k' + r«) The left member of (5) becomes integrating, differentiating. dx di d^x dcp ^ dt ' rq) d}q) IF (6) (7) (8) (9) which are the equations of motion when there is no slipping. The value of 7" is not ec[ual to //IT after the time ty, and to find its value combine the first of (1) and (2) with (9), giving 28 ;484 PROBLEMS. .-. T=0; (10> hence it requires no friction to cause the point of contact to" have no progressive motion. The motion is the same as if the body were in void space under the action of no forces, having a uniform motion botli as to the translation of the centre, and of rotation about the centre. The total amount of slipping will be rep - x = ^oDoh - iMff — ^ tl (11) ; 37. If a sphere, radius 3 feet, weight 20 pounds, rotating ten times per second, be placed on a horizontal plane whose coeffi- cient of friction is ^^ ; how long will it be in coming to a uni- form velocity, how far will it have traveled, how much will it have slipped, what will be the uniform velocity of the centre, and the uniform angular velocity ? 38. If a cylinder have the siame amount of material, diameter, and rate of rotation, as the sphere in the preceding example,^ and placed on the same plane, which will first attain a uniform motion, which will have the greater uniform velocity, and which will have slipped most ? 39. If a cylinder whose altitude equals the diameter of the sphere of example 37, the same amount of material and rate of rotation, be placed on the same plane, which will finally attain the greater uni form rotation ? 40. What must be the coefficient of friction that there be no slipping at the point of contact at the beginning of motion ? 41. Which will first attain a uniform motion, the sphere in example 37, or a sphere of the same material and twice the diameter ? 42. If the body gradually contracts, retaining a constant mass and same form, will it go farther or not before attaining a uni- form velocity ? 43. If a sphere preserves the same circle of contact, but grad- ually contracts laterally, changing to an oblate ellipsoid, will it FRICTION. 435 affect the time of attaining a uniform velocity ? What will be the time if the polar axis becomes half the equatorial diameter ? 44. A rope is stretched round a rough cylindrical surface sub- tending an angle d, the coefficient of friction being ja ; required the force F acting in the direction of the tangent at one end in order that P will be in a state just bordering on motion towards F. Let the tension at any point, a, be t, that adjacent, b, will be ^ + dt, p the normal pressure on the arc per unit of length if it were uniform ; hence, on an element of length, it will hepds, j.i the coefficient of friction ; then dt = j^i ' pds, also pds = Vt* + {t + dt)^ + 2^(^ + dt) cos (n - 6) = tVHi + cos (;r — 6/)), (ultimately) = ^ . 2 COS \\{7r - d) = t '2 sin ^e = idd (ultimately) ; :. dt = f-it . dd. Integrating, But for and for log t = ^e ■\- C. 6 = 0, t = F, e = AB, t = P; .'. F = Fet"^. Prom the relations ftnd pds = tde, we find ds = rde, 4 t P = -. the same as equation {p\ p. 189. 436 PROBLEMS. 45. A shaft having a bearing the entire length is driven by a; pulley at one end, the power being taken at the other. Find the diameter of the shaft at any point for uniform strength. Let w = the weight of the shaft per unit volume, ju = the coefficient of friction, h = the radius at the driving end, x the distance of any cross section from the driving end, and y the radius of that cross section. Then wfATty'^dx = friction of an element, WfXTty^dx = moment of friction. Then Uv^ny^dx + PR = cy\ {a) PR being the moment of the driving power. From {a) wf^Tty^dx = dcyHy, or, letting W}J.7t _ Jo J, y Ax = log - , y and y = Jie-^\ ATTRACTION. ' 46. Assume that two spheres of the same material as the earth, leach one foot in diameter, are reduced in size to a mere point at their centres, and placed one foot from each other, required the time it would take for them to come together ly their mutual attraction, they heing uninfluenced ly any external force. Let E= the mass of the earth, m = the mass of one of the spheres, m' = the mass of the other sphere, R = the radius of the earth, r = the radius of one of the spheres. ATTRACTION. 437 r* = the radius of the other sphere, g = the acceleration due to gravity on the earth, f4 = the acceleration due to the attraction of a sphere of mass unity, upon another sphere of mass unity, the distance between their centres being unity, a = the original distance between the centres of m and m', and X = the distance between their centres at the end of time t. Then from (227) we have which integrated (pp. 33, 34), observing that for ^ = 0, a; = a, and v = 0, and that // in the reference equals (m + m') -^ in this case, gives '=bi^^TO]'4<"-'>'---©'i:' which for the limits gives , _ . r -^^ 1^ If both spheres are of the same density, their masses will be as the cubes of their radii ; or ♦w' = ^^; and we have and if the spheres are equal, as in the problem, we have , , /Ra\i 438 PROBLEMS. If tf = 1 foot, R = 20,850,000, r = 4 foot, g - 32 J, we haye /-_ ^«^,/166,800,000\i = 1,788 seconds, nearly, = 29.8 minutes, nearly. 47. To find the stress in pounds which would be exerted by the mutual action of two such spheres as in example 46, at a dis^ tance of one foot between their centres, ive have, from equations (224) and (235), since m = m', and x=l, l^Eg The mass of the earth is 5| times an equal mass of water. The weight of a cubic foot of water is 62 1 lbs. at the place where g = 32 J, and the volume of the earth is ^7tB\ hence E=bl X 621 X i7tR\ which, substituted above, gives for the stress |7r X 5^ X 62| X 32j 64 X 20,850,000 = .00003471+ lb. 347 or nearly of a pound, a quantity inappreciably small. 48. If the density of the earth at the surface be unity, and at the centre is m, g the force of gravity at the surface, and f thai at any didance x from the centre loithin the earth, what is the law for f, the density increasing uniformly toward the centre.. The density at any distance, x, from the centre will be (m — 1) , \ , i V — — L{r -X) +1. have. and If bence. ATTRACTION. 489 The volume of a sphere of radius a; is 4;r 7?dx. Hence wa Mass = 4;r r H^ ~ ^^ (r - x) -\- llx'i/x, /=4'['^(f-?)-a x=r, /=g, ff = o^, [4(m + r - l)r - 3(m - l)r»] ; _ 4(m 4- r — l)a; ~ 3(m — l)a:' ** "^ ~ 4,m + r — l)r — 3(m — l)r*'^' 49. Considering the earth and moon as uniform todies, t?is mass of the earth 5 J times that of an equal volume of tvater, and its radius 3, 056 miles, the mass of the moon 3^ times that of ajt> equal volume of water, and its radius 1,080 miles, the mean dis- tance between the centres of the earth and moon CO times the radius of the earth, and the acceleration due to gravity at the surface of the earth 32^ feet per second ; required the time in which they would come together by their mutual attraction. When they are in contact the distance between their centres will be 5,036 miles, and we have from Problem 46, But 7M R' 7 Mr' and '-[^?^rJ['"-^''"--e-)T.<" •tiO PKOBLEMS. Substituting numerical values in equation (2), r 3,956 X 237,360 x 5,280 x Slir/ \4 '- Ll93[(3,956)-+-^(l70807]J [(.^^^'^OO x 5,036-(5,036)«^ .237,360 CO.-. yg,)*; = 412,945 seconds. Loomis gives 415,600 seconds as the time required for a par- tide to fall from the moon to the earth, the distance to the moon being the same as that given in this example, which is 2,655 seconds, or nearly 44 minutes more than the time for the earth and the moon to meet. 50. When the earth is in 'perihelion, suppose the sun's mass to he increased hy x times its present value. Required the change in the elements of the terrestrial orMt. The square of the eccentricity of any planetary orbit is (Prob- lem 4, page 189), , ^ FV sin^ 13 V'r^ sin^ /J J. — 4 + in which V is the velocity of the planet at the point whose radius vector is r, /:> the angle between the curve at that point and r, and ^ a measure of the attractive force. In this case a = 90°, and the equation becomes, by reduction, FV , M If now the mass of the sun is increased suddenly x times its present value, pi becomes xjj, and the eccentricity of the new orbit will be Ci (say), hence, FV , XJii ' and ^ FV , ±ei = 1, XJJ. ■; =l{l±e)-l. ATTKACTION. 441 m T^hich Bx must be used when the right member of the equa* tion \s positive, and — e^ when it is negative. \i X < \{l ± e), 61 > 1, or the orbit will be a hyperbola. *' x= 1(1 ± e)y e^ = 1, '' " " parabola. " a; \ ^\^} "^ ^} i , ^i < 1, '• '' " an ellipse. ( < (1 ± e) ) *' x=l±e, ±e^ = '' " *' a circle. " a; > 1 ± e, — 61 > — 1, " " ** an ellipse. In all but the last the given and resulting perihelia coincide ; but in the last the given perihelion will coincide with the new aphelion, cc = | and e = 0.016784, the eccentricity of the earth's orbit, we fiud — ei = —0.32214, or — ^i = 0.34452 ; hence the present perihelion would become the aphelion point of the new orbit, and the new orbit would be an ellipse with the eccentric- ity 0.32214 or 0.34452. The mean distance would become ^^ 0.7436 2(1 + r) of its present distance. The time of rotation about the sun would be ^= ^^/[(rT^)'] "" ^^^* "^^y" = ^^^-^^ '^"y"- 51. Suppose infinitesimal aerolites equally distributed through all space, everywhere movifig equally in all directions with a uni^ form and constant absolute velocity. The aggregate mass inter- cepted in a given time by a given stationary sphere is supposed to be known. Determine the effect upon the eccentricity of a spherical planet of given mass and volume moving in an eccentric orbit all of whose elements are known. (Math, Visitor, July, 1880.) If the velocity of the planet be less than that of the aerolites, the same mass will be intercepted as if the planet was at rest. Consider this case. The change of the elements of the orbit will be due to two causes. Ist. The increase of the mass of the planet will increase 442 PROBLEMS. the attractive force between the sun and the planet. 2d. The aerolites will cause a direct resistance to the motion of the planet. Let M be the mass of the sun, m the initial mass of the . planet, s the distance between them, Jc a constant ; then will the acceleration of one body in reference to the other at the end of time t (the time in the problem being unity) be MM ■\-m' ■¥ mt) hence at distance unity, pt = h{M + m' + mt)y d^ — kmdt. For an elliptical orbit we have, page 190, Vlrl = /^«(1 — e*) = c, a being the semi-transverse axis. Since the changes are small^ consider two quantities only to vary contemporaneously. If $ be constant, we find , hmc ,. da= 7, sr dt ; M\l-e') ' . Icmct 1 therefore. Similarly, if a be constant, we find therefore. 2ejd . Jcmct , Ae = X —====. nearly, ^H- va/< (a/i — c) Hence, the major axis decreases and the eccentricity increases with tlie time, and the amount of change for one revolution may be found by makinc^ t equal to the corresponding time. 2d. The law of resistance is not given. The aerolites being ATTRACTION. 443 infinitesimal, we do not consider the impact as between finite maisses, but they constitute a medium through which the planet moves. Considering the medium as of uniform density, /), and the 'resistance varying with the square of the velocity, the case comes within one discussed by La Place, Mecanique Celeate, (8925, 89-^6). The equations will become for this case. de KDa ,^ — = au ; e /i K being a constant depending upon the mass and form of the planet. If e = ^(l + 2K'a,6)' a, and «, being initial. The major axis and eccentricity both decrease as the vectorial angle increases, and the orbit becomes more nearly circular. The plane of the orbit will not be changed ; and, finally, the longitude of the perihelion will not be changed. — Mecanique Celente (8916). 52. Show that if two bodies revohw about a centre, acted upon by a force proportional to the distance from the centre, and in- dependent of the mass of the attracted body, each will appear to the other to move in a plane, whatever may be their mutual attraction. Take the plane xy in the plane of the central force and of the two bodies at any instant, the origin at the central force and the axis of X passing through the body a, the coordinates of a being x' and 0, and of the other body b, x" and y", d their distance 444 PROBLEMS. apart ; if the central force at unit's distance, ^the force of h on a, and F' that of a on h (according to the Newtonian law F — F') ; X', X", Y', Y", the axial accelerations of a and b. Then X' Y' Mx' + y" . d ' j^X — X = F jj — d ' } X" Y" Mx" - My" - F' F' x" - x' = — d y\ d ' Y" - Y' • X" - X' ~ X f " -X' ' f which gives the direction of the relative accelerations, and which is parallel to the line ab. Hence, whatever be the direc- tions of motion of the two bodies or their absolute velocities, their relative positions {a, b), their relative velocities, and their relative accelerations are parallel to a plane. Solution by Quaternions. — Let p and p^ be the vectors of the two bodies referred to the centre as origin. Let p' = —- , p/ = -^ , p" — -^, p(' — -§- . If il^ is the central force at dt dt dt the unit of distance, and N and JVi the mutual attractions divided by the distance apart, we have p" =-Mp + N{p, - p), a" = — Mpi + N-,(p — a) ; .-. p." - p" = - M{p, -p) + (iv, + ]\r){p - ff). The scalar part gives ^(p-p')(p'-P.')(/3"-A")] = (^+^+-^.)'S(p-A)'(p'-P'0=<V PROBLEMS. 446 which proves that the apparent orbit is a plane. — (Coordinate Geometry, p. 278, eq. (3).) {Math. Visitor, July, 1880.) 53. An elastic string y without weight and of given length, has one end fixed in a perfectly smooth horizontal plane, and the other to a point in the surface of a sphere, the string being un- wound. The sphere is projected on the plane from the fixed point with a linear velocity v, and an angular velocity qi, tcinding tJie string on the circumference of a great circle ; required the elon- . gation of the string when fully stretched, and the subsequent motion of the sphere. Let r = the radius of the sphere, a = the original length of the string, go = the initial angular velocity of the body, v — the initial velocity of the centre of the body, and t^ — the time of winding the slack. Then v^i -f rootx = a ; n. :. t, = and the initial stretched part will be vti = = I (say). Immediately following this time the string will be stretched, and the tension at first diminishes both the linear and angular velocities. Take the origin at the remote end of I for the variable motion. Let m = the mass of the body, s = the space passed over by the centre during time t, 6 = the angular dis- tance passed by the initial radius in the same time, k = the radius of gyration of the body, e = the coefficient of elasticity of the string, A the cross section of the string, and X the elonga- tion produced by the tension T of the string. Then Mariotte's law gives Assume that / is so long compared to rO, that the latter can be neglected, and let B = eA -r- I, then T = B\. 446 PEOBLEMS. The conditions of the problem give dX = ds + rde ; (3> .-. d:'X = d's + rd^e. Also, for motion of the centre, and for the rotary motion, mk'^ = - Tr = - Br\, (4) which two equations in the preceding give Integrating, observing that for X =0, t = 0, and dX-r' dt = v + roo, we have X= — — — &mDt (5) The elongation \ will be a maximum for sin Dt = 1, or ^ = ;r -7- 2i>, for which The time of producing the maximum stretch of the string is independent of the initial motions. When the string returns to its original length \ will again be zero, and sin Dt = 0, or Dt = Tt; .'. t = ^. All the circumstances of the variable motion may be deter- ^ mined by integrating equations (3) and (4). Integrating after ds substituting from equation (5), observing that for t = 0, _~ = v, « = 0, ~— 00^ and ^ = 0, we have, if we put i^for eA{v + roo) PROBLEMS. 447^ g- = ^[C08/>^-l] + tf, (6) F s-j^ [sin Dt - Dt] + vt, (7) ^ = ^p[C0Si?/f-l] + 07, (8) ^ = -^ p [sin Z)i? - Z>if] + (w^. (9) For the maximum of (5) d\ ^ dt = 0, which in (2) gives ^_ _ dd dt ~ ~'^di* which combined with (6) and (8) gives cos 7)^ — 1 = — 1 ; .*. Dt = \7r BA before found, and serves as a check upon the work. The relation ds — — rd6 shows that the direction of one of the motions changes sign. At the point where the linear ds motion is reversed, ^ =0, and for this we have ^ = icos-(l--|,); and if the direction of rotation is reversed, ^ = 0, and (8) gives 4=. 1 , / GDk'\ from which it appears that if v < k'^oo -i- r the motion of the centre will be reversed, but otherwise the angular motion will be reversed. The value of tt in the former case will be less than -jz . Both motions will change at the instant of greatest elon- gation if rv = Jc^QO, If the values of t^ and t^ are both less than n ^ D, one motion will change sign before the instant of greatest elongation and the other after ; otherwise only one will change sign. To find 448 PROBLEMS. the total variable moyement, make Dt = n, and (7) and (9) give e (-4)^ If (6) reduces to zero when Bt = tt, the body would rest at the moment the string regains its original length, and F= |^;, but it would still have an angular velocity of oo + {rv -^- k'^), as shown by (8). Similarly, if the rotary motion is destroyed at that instant, the linear velocity will he v + (kroo — r), and will continue uniform. It may be shown that the kinetic energy, of the moving body at the end of the variable motion is the same SiS, at the beginning. It has not been attempted to solve the general case represented by equation (1). It is evidently intricate. — (Problem and solu- tion by the author in The Analyst, Jan., 1882.) 54. Find the minimum eccentricity of an ellipse capable of resting m equilibrium on a perfectly rough inclined plane, in clination y^. The centre of the ellipse will be vertically over the point of support, and since the plane is tangent to the ellipse, a vertical through the point of support and a parallel to the plane through the centre will make conjugate diameters. Hence the acute angle of the conjugate diameters is 90° — y5. At the point bordering on motion, the potential energy is a maximum, and the major axis will bisect the acute angle of the conjugate diameters ; hence the positive angles made by the conjugate diameters with the major axis will be ^ = 45° — Jy^, 6' = 135° + Jy5. The condition for conjugate diameters is a^ sin sin /9' + b^ cos cos d' = 0, which, by substituting the preceding values, gives a' sin^(45° - i/i) - ¥ cos^(45° - i/3) = ; FLUIDS. 441 which, reduced, gives {Math. Visitor, Jan., 1879.) FLUIDS. 55. A sphere 4 inches in diameter, specific gravity 0.2, w placed \Ofeet under water. If left free to move, what will be its velocity at the surface of the water, and what will he the maxi" mum height it will attain. Let r = I — radius of the sphere, p= \= its specific gravity, h = 10 feet, V the velocity acquired in ascending a distance af, Fthe velocity at the surface of the water, k = the resistance. For the motion of the sphere, vdv Putting^^--lj = /, , vdv dx — - !>=[ kv^' vdv whence. g -kv""' = ^^—77—^ nearly. The i-equired height is F« 1-p 450 TROBLEMS. The value of h reduced from Newton's Principia, Book %.^ 3 Prop. 38, is — — , which gives and F=[Vr^(l-p)]* = |v^ A = — = 4:f inches nearly. {Math, Visitor, Jan., 1879.) 66. The first of two cashs contains a gallons of tvine, and the second b gallons of water j c gallons were draum from the second cash, and then c gallons were drawn from the first cask and poured into the second, and the deficiency in the first supplied hy c gallons of water ; c gallons were then drawn from the first cash, and c gallons from the second, and poured into the first, and the deficiency in the second cash supplied hy c gallons of wine. Re- quired the quantity of wine in each cask after n such operations as that described above. Let u^ and v„ represent the wine in the first and second casks respectively at the end of the nth operation ; the quantities of wine in each cask at the successive stages of the {n + l)th opera- tion are u., v.;(l- 0«., (l -!-)«'.; (l - '-) n„ {^-'^v.+l n, V Whence FLUIDS. -=(i-0-<-(^-O— 451 (2) Also (3) (4) Eliminating v^ from (1) and (2), Eliminating v,+i from (3) and (5), -.-[('-0"-i-(-i)>--('-:-)'('-O'- Let r„ ra, be the roots of the equation ^-[('-:-y-^(-^)>^(-:-)'(-iy-- The solution of (6) is, (Hymer's Finite Differences, pp. 54, 65), «.=c;rr+c;rj+ -(-3-:i-(-i)H'-0('-i)l Let the last expression — S. From (1), u^ = a, a 452 PEOBLEMS. whence ^'- — ^^r=7^ — ' C - ^'^^ - S) -Ui + S n - r, ' and Eliminating w„ from (1) and (2), and w„,i from this and (4), '--[('-:-)'*'iK'-0']--('-:-)'(-r)'- whence From (2), t;o = 0, v^ = c(^^Y)> therefore (i/«if7i. Visitor, Jan., 1880.) 67. J servant draws off a gallon a day for 20 days, from a cask holding 10 gallons of wine, adding each time a gidlon of water to the cask, he then draws off 20 gallons more, adding as taken, a gallon of wine to the cask for each gallon drawn. How much water remains in the cask 9 Put 10 gallons = a. 1 gallon = 5. 20 = t, and let u^ = the number of gallons of water in the cask at the end of the orth operation. Then we have \~~^)'^^' + J = w«+i, (1) FLUIDS. 463 an equation in Finite Differences. Integrating (1), a — h^ ^'^Kv^y^-^- ^^> When x = 0, w, = ; .*. C=— 0, fa' -{a- hy\ by the first condition. Let V, = the quantity of water in the cask at the end of the ^h operation under the second condition ; then. Integi-ating, - =<'—')•• When / = 0, Vo = c; ••• 0==Ci, Vt = c C^—^ ) > fa - by /fl- - (« - m When x = t = \0, a -JO, b=l, v, = 10 ( 9^^ - .9*'). (7 'he Wittenherger, Jan., 1880.) 58. A piston, weight w, is dropped into the end of a vertical cylinder filled with air, length I ; hoto far toill the piston descend, assuminq no friction nor escape of air, nor heat from the com' pressed air f 464 PROBLEMS. There being no escape of heat, the law of pressure will be ex- pressed by pv'^ = constant =j)'v''^, (1) where p' = the initial pressure of the atmosphere = 15 lbs. nearly ; v' = volume of the cylinder = al, if a is the area of the base ; p = the pressure within the cylinder when the weight has descended a distance x ; a(l — x) = the volume when the press- ure is ^ ; k = 1.408 ; then, from (1), and pT gdf ^ (I — xy Integrating, observing that tor x — 0, v = 0, v'=il-h^ l^\2nr 4- ^1^1^ n{l -xY-'-l^ \ w J^ ^ w{k~l)\ {l-xf--^ / At the end of the downward movement v = ; therefore x(w + pa){l - xf-^ -V- ^^^ (/ - a:)*-i = f^y from which x may be found by trial after numerical values have been substituted for the known quantities. 59. If each of n vessels closely connected in circuit contains a different liquid, each (\ gallons, and the liquids circulate by Jloiv- ing uniformly in one direction at the rate of a gallons per min- ute, mixing uniformly, how much of each liquid luill there he in any one of thp vessels at the end of the time t. Let the vessels be numbered in the natural order 1, 2, 3 ... /i ; let x denote any particular liquid and a;,, x^, x-^, . . . x^, the quantity of it in vessels 1, 2, etc, respectively at time t, adt = the FLUIDS. 45fi( ^amount flowing out (and in) each instant, of which — adt. '^adt ^n adty will be of the liquid denoted by x. In vessel X X 1, - adt flows out, and — ^ adt flows in from the nth. vessel, and the difference will be an element of the decrease of the x liquid ; («) dx. = — adt adt. q q Letting — = r, and dividing by dt, we have or rdxi ~df rdxi ~df = Xx — Xi= — Xn. Similarly for the others, rdx^ "df Xa Xl rdXi _ di ~ ""'' ~ "^^ rdx^ ~df — x„= — iC._, Differentiating n — 1 times, we have (*) rd'Xi df^- d^-^Xi = - d-'x, dt'-' rd--% dt"-' d''-*x, dt^-* = - d-'x^ dt"-* rdx^ dt -Xi= - -X, («) 466 PROBLEMS. Similarly, dV'' dV'-^ ~~ ~ dF^' dt''-' dt"-^ "" 'di'^' rdxt and finally. dr~^'~ ~^' ' dt" df"-' ~ 'df-^ * rdxr,-x _ Multiplying the successive equations in (c) by the successive terms of the development of the binomial (r —1)""', and add- ing results, we have ^W dt''-' "•" [2 dF^^ '-^ ~df "^ ^' The second member of this equation is one degree less than the first, and since this is true for all values of n, it will be FLUIDS. 457 true if we reduce the exponeot and subscript still further by 1. This process ultimately gives for the second member. rdx^ hence we have =^^r'^'^'='^'^^' r^/% nr*-^d^-% n{n — 1) r"-'c?*-*a;i dt^ dt^-' "^ |2; dt"^^^ • • • n(n — 1) r'd'x^ nrdxi _ Adding ^ 1 to both members, the characteristic equation becomes (Price's Infinitesimal Calculus, Vol. II., p. 634), f^fi^ — wr"-'/?"-* ± nrfi :f 1 = q: 1, or (r/J-l)-=q:l; ^ = -T ' which may be written p=\±ve:^. (rf) where the exponent n, of — 1, is simply for the purpose of determining the sign of this term. If n be even (— 1)" will be + 1, and there will be two real roots + 1 and — 1 ; if w be odd there will be one real root of \/ — \=. — 1. The other roots will be imaginary and in pairs. The number of values of P will equal the degree of the equation, one value being zero, and letting these values be i„ ^„ 63 . . . J._i, the integral becomes (Price's Infinitesimal Calculus, Vol. IL, eq. (107), p. 637), X, = C,d>^^ + C^«' 4- + C:^«S («) and similar expressions for a;,, x^, etc. 458 PROBLEMS. To find the constants of integration we have for ^ ^ 0, x^ — q. dXy , = a, from (a) since, initially, 2;„ = 0, ^r\ _ /a dxi a dx,\~[ _ «' _ /^ _ «' . di" Jt=o~ \q~dt~q Wy J,=o~ q ~q' ~d7'jt=.o q .'. q= C, -h C,+ C, T, , etc. (/) = b\G, + hlC^ h\a and finally _ = JnQ+5» W. from which the constants of integration can be found As a special case let 7i = 3, then from [d), ^ = **' 5 <^ - ^^' ^""^ i (^ + ^^^^- We then have from (e), and similarly. Xi= A^ + A^e 2-(3-V-3)« + A^e ,(8+>/-3)< ^(3-v/-3)< §'«+^-3)« (g) x^ = B-^ + B^e^'^ + The sum of these, or ic, + ^2 ■¥ x^^q at any time t. FLUIDS. 469 Also by the process shown in (/), we have f or ^ = 0, a^ = q, -di = ''' W^q'^ ^'^^^ W = ~^' W=-Y' ^ = ^' ~dt ~ ' dt* ~ q ' Hence, to find the constants in equation (^), we have, for Xi , q = C, +C,+ C^, « = ^(3 -V^3)C;+ ^(3 + V- 3)C« ^' = |. (3 -A/^3r(7. + |-, (3 + V- 3)«Ci 5 fora^ = ^, + ^, + At, - T = ^ ^^ " '^~'^'^' "" ¥ ^^ ^ V~3)M. ; and for x,, = 5, + 5, + £„ = ^ (3 - V- 3)fi, + J (3 + V- 3)5„ ^ = ^ (3 - V- 3)'5. + ^ (3 + V^)-5, These give ^, = C, = Ca = i^. 5, = ^^, B, = - iq(l - V^3), B,= - iq{l + \/=^). 460 These in (g) give PROBLEMS. Sat / at /—^ at/ — -\"~j = iq(l + ^ 2e 2? cos — -"" ) ; ^2=i^r2 - Sat COS y +V3 sm~— ) ; ^7 ^(Z /J i^ ^ D 3a« 23- COS V'Sal Vasm-'V^^^)]. In a similar manner, if ^ is tbc liquid in tlie second vessel, the quantity of it \n vessci ^j ii'c tiic end of time t will be ?/i = iCi above ; and similany for tiio others. 60. 2 find the velocity of an ice boat. Let AB represent the track of the boat, BD the position of the sail, 6 — DBA and WG the direction of the wind, which we will assume to be normal to the sail. When the boat ad- vances to C, the posi- tion of the sail will be CE. If V be the ve- locity of the boat, pro- portional to BC, and V the velocity of the wind relatively to the earth, then will v — F sin ^ be the velocity relatively to the sail, since the wind passing' any point as B must travel a distance BIT before coming in contact with the sail. The pressure of the wind is assumed to vary as the square of the velocity relatively to the surface pressed, and if Jf be the mass of the boat and sail, B a constant depending PROBLEMS. 461 upon the size of the sail and the unit of velocity, and neglecting all resistances, we have (§) (V ds\ Adt, Barn^e where A = — jr^ — . Integrating, making F = for ^ = 0, ^^ - V — sin 6 ^ dt~ ~ sin () ' Avt from which it will be seen that V increases as t increases in- definitely, the limit of V being sin 6 and this increases as decreases, from which it appears that, according to the above hypothesis, the boat might attain an immense velocity for a small velocity of the wind. The smaller the angle of the sail with the track of the boat, the greater the ultimate velocity of the latter, there being no resistances. But the resistances may be considerable. The coefficient of friction on the ice may possibly be as low as 0.04. Should the air move away from behind the sail with the velocity of the wind, or even with the velocity Fsin 0, no resistance would be offered by the air ; but such will not be the case. The ma^t, sail, and other parts will be opposed by the resistance of the air as tliey move through it, but it is difficult to determine the exact amount. General Tower, in Van Nostrand's Engineering Magazine, January, 1880, pp. 83, 84, according to an exampie of probable conditions, concluded that the maximum velocity of an ice boat might exceed twice that of the wind. 462 PROBLEMS. 61. Determine the path of a rotating tody projected into a resisting medium, A general solution of this problem has not been obtained, but certain qualitative results may be determined without finding the quantitative. Let the body be a sphere rotating about a vertical axis, the centre moving in a hori- zontal plane. Let od be the angu- lar velocity, and v the velocity of the centre ; then will OK, the distance of the spontaneous axis K from the centre be, equation (199), OK The combined motion of rotation and translation of the body at any instant being considered the same as that of the entire body rotating an infinitesimal amount about the axis K, the quadrants A and D will move with equal velocities exceeding those of B and G, The resistance of the medium to a body moving normally against it will vary as some power of the velocity of the body, and in this case may be considered as pro- portional to the same. The quadrant A moves against the medium with a greater velocity than B, hence the pressure on that quadrant will be greater, while the velocity of the quadrant D moving away from the medium, exceeds that of C. There- fore the pressure of the medium will be greatest on quadrant A, next B, then C, and least on D, The resultant of these press- ures will not be zero, and generally not parallel to OG, Let R be the resultant, the component of v/hich, parallel to OG, will be the pressure directly opposing the motion, and ab, normal to OG, the pressure which will deflect it from its initial direction. Neglecting friction, the path will be a curve convex towards the quadrant of greatest pressure, and will be more nearly a right line as K is more remote, or the more v exceeds od. Still, neglecting friction, the rotary motion will be constant, while the velocity of the centre will be diminished ; hence the curva- ture of the path will increase with the distance traveled The friction between the medium and body tends directly to PROBLEMS. 463 diminish the rotation, but if the sum of the components of the frictional resistances resolved in reference to two rectangular planes be not zero, there will be a resultant. The friction will be greatest where the pressure is greatest, as at ^, and act tan- gentially to the surface. Lot F be the resultant ; it will be equal to a couple, and an equal parallel force at the centre, the former of which reduces the rotation ; and of the latter, that component which is parallel to 00 directly opposes the motion, and that which is normal to OG tends to deflect the path in the direction OE, opposite to that produced by pressure only. If the body be comparatively smooth and the medium rare, the friction will be only a fractional part of the pressure, and the resultant friction will be only a very small part of the entire friction, in which case the direction of the curvature of the path will be de- termined by the YQ^nMant pressure, but the amount of curvature will be diminished by the friction. This case is illustrated by a rotating sphere projected into air. But if the body be rough and the medium dense, frictioual resistance might exceed the pressure, in which case the direc- tion of curvature would be determined by the resultant friction, the amount being modified by pressure. This could be illus- trated by a wheel with flat vanes rotating about its axis, placed vertical and pushed along in water. Or still more strikincfly, if a rongh cylinder rotating about a vertical axis be pushed into a bank of earth, the tendency to a lateral motion might be almost entirely dependent upon the friction. APPENDIX II. THE POTENTIAL. THE POTENTIAL. The term ''potential function," or simply the potentialy as used by Gauss and subsequent writers, is applied to a certain expression appearing in certain investigations involving forces depending upon some function of the distance between the bodies. It is of value in the higher investigations of the theory of attraction, hydromechanics, electricity, magnetism, and heat. Before defining it definitely, take an example. Let m and m' be the masses respectively of two particles, the place of m being a:, y, z, of m', x', y z', the distance between them r, and f(r) the law of tlie mutual attraction or repulsion. Then will the stress between them be P=^mm'f{r), (a) — being attractive and + repulsive. The axial components will be P cos ^^ :=: X= mm'f{r) — "^1?, Y= mmf(r) y~~JL , Z=mm'f(r)^~-^', {b) and for the distance between them, r" = (a:' - a-)* + (//' - y)' + {z' - z)' ; (c) from which we have for the partial differential coeflficients, x, y, fi, being considered fixed, rdr = {x' — x)dx\ rdr = {y' — y)dy\ rdr = (z' — z)dz. (d) heifer) be considered as the differential coeflBcient of some other function of r, so that 467 468 APPENDIX n. then will equations (d) and (e) reduce (b) to ^ , dF(r) ^ , dF(r) „ , dF(r) If m^, nis, etc., be the masses of other particles distant r„ r^ etc., from m', then we have, m being typical of m^, m^y etc., ^ ,d2mF(r) ^ ,d:^mF(r) ^ d^mFir) ... ^=^ dx' ' ^=^ dy' ^ ^='' -~W^'^^ Let then we have X=m I -7- MX '). y-'Q- '-<-£)■■ « the parentheses indicating partial differential coefficients. The function V is called the potential, and may be defined by inter- preting equation {g). When found, the axial components of the stress appear as partial differential coefficients of V regarded as a function of the coordinates of the particle. It is rarely used in this general form, but is confined to the cases where the law of the force is that of the inverse squares — the law most common in nature. Let the system of particles m be continuous, form- ing a solid, and the force attractive, then we have and if ^ be the density at the place x, y, z, then im = ddxdydz ; ddxdydz ^ ^^ that is, The potential is sum of the quotients of all the element^ ary masses divided hy their distance from the attracted particle. THE POTENTIAL. 469 Again, if the mass of the attracted particle be unity, and that of the attracting particle be m, then at any distance r we have the stress. and hence an element of the work done upon the unit mass at that point in being moved over the space dr will be -\ J^^r=-; (J) and similarly for any number of particles forming a continuous body ; which is the same form as that above, hence, also. The potential is the energy acquired by a unit mass infaU- ing from infinity under the attraction of a given body to a dis- tance r. Again, rw may represent any quantity of action, either attrac- tive or repulsive, as in magnetism or electricity, in which the law of action is that of the inverse squares and product of their quantities. From equation (c) we have the partial derivative (dx') a:' — a;' and from (J), considering F as a function of r. (dV) = I 1 1 ~ ^dxdydzdr /^^\_ f f f S{x' -x)dxdydz 470 APPENDIX TI. similarly. (£f)=ifj['^--a**-^ and adding fd'V\ /d'V\ fd'V\ ^ which theorem was discovered by La Place. It is not general, however, for it is found to fail when the particle is one of the particles of the attracting mass ; but it is correct when the particle attracted is external to the attracting body. Examples. 1. To find the potential of a slender uniform rod, length a, density S, and section s, upon an external particle m. Take the origin at one end of the rod, x, along the rod, and x', y', the position of the particle m'. Then 17 r ^^^^ ^ 1 a-x'+{y"-\-{x'-ay)^ . , Jo a/(«' - xf + y" ^ -x' + {x" + xj ^ ^ Hence the attractive forces parallel and normal to the rod will be respectively, \dx J [_^y'^ + (« - xj Vy" + ic'd \dy J y L^/y'^ + {a- xJ ^/y'^ + a;'«J 2. To find the potential of a thin, homogeneous, spherical shell upon an external particle. THE POTENTIAL. 471 Lefc a = radius, d = density, da = thickness, p = distance of pai'ticle from the centre of the shell. Using polar coordinates, origin at the centre of the shell, (0, 0, /j) the i)lacc of m\ = polar distance of element of m from where p pierces the shell, cp — longitude, initial at any point, then dm = 6 da - a sin Odqj • add, r' = rt ' — 2ap cos 6 -t p' ; JT j,^ P f^" sin fidddcp J J (« — ^ap cos 6 + p")i ' 27r6ada / „ ^ n , iVl"" = ( a' — 2ap cos (9 + p ) , -{a- p)j, (n) -(p-«)). (0) or. 27r6ada 27tSada a + p For an external particle p > «, hence the last equation is the correct form for this case, and the former gives the potential for a particle internal to the shell. For the latter. A:7t6a^da _ m p p and for a concentric shell of finite thickness, •*• \dp) - p" which is the force along the line p ; hence, TTie attraction of a spherical, homogeneous shell upon an 472 APPENDIX n. exterior particle is the same as if the mass of the shell were cort" densed into a particle at the centre. Equation (n) becomes V = 4:7tSada ; •••(S=». hence. The attraction of a spherical homogeneous shell upon a particle within it is zero. ""^^ ^OOK IS DUE ON THeTas^aTE STAMPED BELOW -- L'^'Js's^is.^'^^'^ 25 CENTS THIS BOOK ON THE DAT. ;^"-"''^ ^° """"N WILL INCREASE T0%nJS °''^- ™^ PENALTY OAV AND TO Joo ON T-"" ^^ '^°"''™ OVERDUE. " ™^ SEVENTH DAY MAfr la i9a& SEP 26 1939 OCT 26 1933 kl-! 131937 J4/V 2'T> J944 ^^^ n 1938 LD 21-50m-8,-33